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Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Solution of Trigonometric Equations 

A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and can be classified as :

(1) Principal solution

(2) General solution.

(1) Principal solutions : The solutions of trigonometric equation which lie in the interval [0, 2π) are called principal solutions. 

(1) General solutions :  The solutions expressing all the values which would satisfy the given equation, and it is expressed in a generalized form in terms of ‘n’.

General Solutions of Trigonometric EquationsGeneral Solutions of Trigonometric Equations

Solved Examples:

Ex.1 Find the Principal solutions of the equation sinx = 1/2.

Sol. 

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

sin x = 1/2

there exists two values 

i.e.  π/6 and  5π/6 which lie in [0,2π) and whose sine is 1/2

Principal solutions of the equation sinx = 1/2 are π/6 and  5π/6

Ex.2 Solve  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.3 Solve tan θ = 2 

Sol.  tan θ = 2 ....(1)

Let 2 tan α ⇒  tan θ = tan α 

 ⇒ θ= n π  + α, where α = tan–1 (2), n ∈ I

Ex.4 Solve Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.5 Solve Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & AdvancedTrigonometric Equations | Mathematics (Maths) for JEE Main & Advanced        

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Question for Trigonometric Equations
Try yourself:
Find the principal solutions of the equation cos(x) = -1/2.
View Solution

Solutions of equations by factorising 

Ex.6 Solve the equation sin3 x cos x -sin x cos3x = 1/4.

Sol.
The equation can be written as 4 sin x cos x (sinx – cos2 x) = 1,

= –2 sin 2x cos 2x = –sin 4x = 1

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.7 Find the general solution of the equation

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.8  Find the general solution of the equation sin3x(1 + cot x) + cos3x(1 + tan x) = cos 2x.

Sol.

sin2x(cos x + sin x) + cos2x (cos x + sin x) = cos 2x

(cos x + sin x)(cos2x + sin2x) = (cos x + sin x)(cos x – sin x)

either cos x + sin x = 0 ....(1) or

cos x - sin x = 1 ....(2)

from (1) tan x = – 1 or 1 – sin 2x = 1 ⇒ sin 2x = 0

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

If sin 2x = 0 ⇒ 2x = nπ ⇒  x = nπ/2 this is to be rejected because of the tan x or cot x will not be defined so x = (nπ - π/4), n ∈ I

Ex.9Find the solutions of the equation, 
Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

2 sin2x = 1 + cos x ; 2 cos2x + cos x – 1 = 0

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Question for Trigonometric Equations
Try yourself:
Find the solutions of the equation 3sin x cos x = 2sin x.
View Solution

Solutions of equations reducible to quadratic equations 

Ex.10  Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. The given equation makes no sense when cos x = 0; therefore we can suppose that cos x ≠ 0. Noting that the right-hand member of the equation is equal to Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and dividing both members by Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

⇒ (tanx – 3) (tan x + 1) = 0

⇒  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.11 Find the general solution set of the equation logtan x(2 + 4 cos2x) = 2.

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.12 The equation cos2x -sin x + a = 0 has roots when x Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced find 'a'.

Sol. 1 – sin2x – sin x + a = 0 =  sin2x + sin x – (a + 1) = 0 (let sin x = t)

 t2 + t - (a + 1) = 0,

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.13  Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Using the identity (sin2 x + cos2 x)2 = 1 we get sin4 x + cos4 x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.14  Find all solutions of the equation (tan2 x -1)-1 = 1 + cos 2x, which satisfy the inequality Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Let us reduce the initial trigonometric equation to the form Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

The following values of x are solutions of this equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

By the hypothesis, we must choose those values of x which satisfy the inequalities

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.15 Determine all the values of a for which the equation sin4 x -2 cos2 x + a2 = 0 is solvable. Find the solutions.

Sol. Applying the formula sin4 x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced, cos2 x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and putting cos 2x = t

we rewrite the given equation in the form t2 -6t + 4a2 -3 = 0 ........(1)

The original equation has solutions for a given value of a if and only if, for his value of a, the roots t1 and t2 of the equation (1) are real and at least one of these roots does not exceed unity in its absolute value.

Solving equation (1), we find t1 = 3 -2 Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced, t2 = 3 + 2 Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced.

Hence the rotos of equation (1) are real if Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

If condition (2) is fulfilled, then t2 > 1 and, therefore, this root can be dicarded. Thus, the problem is reduced to finding the values of a satisfying condition (2), for which Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

From (3) we find Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Since the inequality Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced is fulfilled for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced the system of inequalities (4) is reduced to the inequality Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Thus, the original equation is solvable if Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and its solutions are

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Question for Trigonometric Equations
Try yourself:
Which of the following values of a makes the equation sin4 x -2 cos2 x + a2 = 0 solvable?
View Solution

Solving equations by introducing an Auxiliary argument 

Ex. 16 Solve sin x + cos x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced sin x + cos x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced              ....(i)           Here a = 1, b = 1

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.17  Solve the equation cos 7x -sin 5x = Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced (cos 5x -sin 7x).

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.18  Solve the equation 2 sin 17x + Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced cos 5x + sin 5x = 0

Sol. Dividing both sides of the equation by 2, we reduce it to the form Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

whence we obtain  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.19  Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Using the formula for the sum of cubes of two members we transform the left-hand side of the equation in the following way : Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Hence, the original equation takes the form Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

The expression in the first brackets is different from zero for all x. Therefore it is sufficient to consider the equation sin x + cos x -1 = 0. The latter is reduced to the form

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Question for Trigonometric Equations
Try yourself:
Solve the equation sin 3x + cos x = 1.
View Solution

Solving equations by Transforming a sum of Trigonometric functions into a product 

Ex.20 Solve cos 3x + sin 2x -sin 4x = 0

Sol. 

cos 3x + sin 2x – sin 4 x = 0 ⇒  cos 3x + 2 cos 3x . sin (–x) = 0 ⇒  cos 3x – 2 cos x . sin x = 0

⇒  cos 3x (1 – 2 sin x) = 0 ⇒  cos 3x = 0 or 1 – 2 sin x = 0

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

solution of given equation is Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Solving equations by transforming a product of trigonometric functions into a sum 

Ex.21 Solve sin 5x . cos 3x = sin 6x . cos 2x

Sol.  

sin 5x . cos 3x = sin 6x . cos 2x

⇒  2sin 5x . cos 3x = 2sin 6x . cos 2x

⇒  sin 8x + sin 2x = sin 8x + sin 4x

⇒  sin 4x – sin 2x = 0

⇒  2 sin 2x . cos 2x – sin 2x = 0

⇒  sin 2x (2 cos 2x – 1) = 0

⇒  sin 2x = 0 or 2 cos 2x – 1 = 0 ⇒  2x = nπ, n ∈ I or cos2x = 1/2

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced  

Question for Trigonometric Equations
Try yourself:
Solve the equation cos 4x + sin x - cos 2x = 0
View Solution

Solving equations by a change of variable

(i)  Equations of the form  P (sin x ± cos x , sin x . cos x) = 0  ,  where  P (y , z) is  a  polynomial , can be solved by the change cos x ± sin x = t ⇒ 1 ± 2 sin x . cos x = t2.

(ii)  Equations  of  the form  of   a . sin x + b . cos x + d = 0 ,  where  a , b & d are  real  numbers  & a, b ≠ 0  can be solved by changing  sin x & cos x into their corresponding tangent of half the angle.

(iii) Many  equations  can  be  solved  by  introducing  a  new variable .  eg.  the equation sin4 2 x + cos4 2 x = sin 2 x . cos 2 x  changes to

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.22   Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.  

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.23  Solve the equation sin 2x -12 (sin x -cos x) + 12 = 0

 Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced


Ex.24 Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Ex.25 Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Ex.26     Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.27 Solve 3 cos x + 4 sin x = 5

Sol. 

3 cos x + 4 sin x = 5

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 equation (i) becomes  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Question for Trigonometric Equations
Try yourself:
Which of the following types of equations can be solved by changing sin x and cos x into their corresponding tangent of half the angle?
View Solution

Solving equations with the use of the Boundness of the functions sin x & cos x 

Ex.28  Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol.

The equation makes no sense for x = π/2 + kπ and for x = -π/4 + kπ. For all the other values of x it is equivalent to the equation

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

After simple transformations we obtain sin x (3 + sin 2x + cos 2x) = 0.

It is obvious that the equation sin 2x + cos 2x + 3 = 0 has no solution, and therefore, the original equation is reduced to the equation sin x = 0  ⇒  x = kπ

Ex.29  Solve the equation (sin x + cos x) √2 = tan x + cot x.

Sol.

Let us transform the equation to the form 

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced  ....(1)

We have |sin a| ≤ 1, and therefore (1) holds

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

or  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

But the first two equations have no roots in common while the second two equations have the common roots x = π/4 + 2k π . Consequently the roots of the given equation are x = π/4 + 2k π

Ex.30 Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. 

Obviously no solution is possible if π/2 < x < 2π as LHS < 2.

If 0 < x < π/2 , then LHS = sin2n – 1 x + 2 cos2n – 1 x < sinx + 2 cosx = 1 + cos2 x < 2 when n ∈ N – {1}.

Obviously, a solution exists only when x = 0 ⇒  The general solution is x = 2mπ, m ∈ I.

When n = 1

sin x + 2 cos x = 2 

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Ex.31 Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Sol.Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

since square of the cosine of any argument doesn't exceed 1, the given equation holds true if and only if we have, simultaneously Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

so, equation (3) has no solution for k ≠ 0 for k = 0

sin x + √2 cosx = 0 or √2 sinx - sinx - √2 or, sinx = -1/√2 , √2

but sin x = √2 is not possible. so only solution to the equation (1) is

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced equation (2) becomes an identity but  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced doesn’t satisfy equation (2) so, solution to the original equation  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

 

Ex.32 Find the general solution of the equation, sin 3x + cos 4x - 4 sin 7x = cos 10x + sin 17x.

Sol. 

(sin 17x - sin 3x) - cos 10x - cos 4x + 4 sin 7x = 0 ⇒   2 cos 10x  sin 7x + 2 sin 7x  sin 3x + 4 sin 7x=0

⇒  sin 7x (cos 10x - sin 3x + 2) = 0 

 Hence sin 7x = 0 ⇒ x = nπ/7, n ∈ I

or  cos 10x - sin 3x + 2 = 0 ⇒ cos 10x = - 1    and    sin 3x = 1  given x = (4n + 1) π/6

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Those starred also satisfy  cos 10x = - 1 ,  the general term of which is

x = 3 (4k - 1) π/6 , k E I , Hence x = nπ/7 or 3(4k + 1) π/6 where, n, k ∈  I

Question for Trigonometric Equations
Try yourself:Which of the following equations has no solution?
View Solution

I. Simultaneous equations 

Ex.33Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Transform the system to the Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Adding together the equations of system (1) and subtracting the first equation form the second we obtain the system

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced .....(2)

The first equation of system (2) can be rewritten as Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

If sin (x – y) = 0, then x – y = kπ. But from the second equation of system (2) we find    cos (x – y) = –1, x – y = (2n + 1)π.

Consequently, in this case we have an infinitude of solutions : x – y (2n + 1) π.

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced then 3x = y = 2kπ. But x – y = (2n + 1) π

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Ex.34 Solve the system of equations Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. Adding up the equations of the system, we arrive at an equation

sin x sin y + cos x cos y = √3/2 ⇔ cos (x-y) = √3/2

Subtracting the first equation of the system from the second. we arrive at an equation

cos x cosu y - sin x sin y = √3/2 ⇔ cos (x-y) = 0

Thus the initial system is equivalent to the system  Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

⇔ n, k ∈ Z, cos (x + y) = 0, 

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Miscellaneous Questions 

Ex.35  Solve the equation 2 cot 2x – 3 cot 3x = tan 2x

Sol. The give equation can be rewritten in the form Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Note that this equation has sense if the condition sin 2x ≠0, sin3x ≠0, cos2x ≠0 holds. For the values of x satisfying this condition we have 3 sin x cos 2x = sin 3x. Transforming the last equation we obtain Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and thus arrive at the equation 2 sin3 x = 0, which is equivalent to the equation sin x = 0. Hence, due to the above note, the original equation has no solutions.

Ex.36  Solve the equation Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Sol. The right-hand side of the equation is not determined for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced, because for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advancedthe function Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced is not defined, for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced the function tan x/2 is not defined and for

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced the denominator of the right member of the right member vanishes. For Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced we have
Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced.

Hence, for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced (where k and m are arbitrary integers) the right member of the equation is equal to -2 sin x cos x.

The left member of the equation has no sense for Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced and for all the other values of x it is equal to -tan x because

Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced

Thus, Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced then the original equation is reduced to the form tan x = 2 sin x cos x.

This equation has the roots Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced It follows that the original equation has no roots.

The document Trigonometric Equations | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on Trigonometric Equations - Mathematics (Maths) for JEE Main & Advanced

1. What are the common methods to solve trigonometric equations?
Ans. Common methods to solve trigonometric equations include factorising the equations, reducing them to quadratic forms, introducing auxiliary arguments, transforming sums of trigonometric functions into products, transforming products into sums, changing variables, and utilizing the boundedness of sine and cosine functions.
2. How can I solve a trigonometric equation by factorising?
Ans. To solve a trigonometric equation by factorising, you first express the equation in a factored form, such as \( a \sin x + b \cos x = 0 \). Then, set each factor equal to zero and solve for the values of \( x \) that satisfy the equation.
3. What does it mean to reduce a trigonometric equation to a quadratic form?
Ans. Reducing a trigonometric equation to a quadratic form involves rewriting the equation in terms of a single trigonometric function, such as \( \sin^2 x \) or \( \cos^2 x \), allowing it to be expressed as a quadratic equation. This can be solved using standard methods for quadratic equations.
4. What is an auxiliary argument in trigonometric equations?
Ans. An auxiliary argument is a technique used to simplify trigonometric equations by expressing them in terms of a single variable or using identities to combine multiple trigonometric functions into one. This often helps to find the solutions more easily.
5. How can the boundedness of sine and cosine functions help in solving equations?
Ans. The boundedness of sine and cosine functions, which range between -1 and 1, can help determine the feasibility of solutions in trigonometric equations. For instance, if an equation yields a value outside this range, it indicates that there are no real solutions.
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