Test: Trigonometric Equations (3 August) - JEE MCQ

General solution of some standard trigonometric equations:

Calculation:
Given: cos x = 1
⇒ cos x = cos 0
As we know, If cos θ = cos α then θ = 2nπ ± α
Therefore, x = 2nπ ± 0 = 2nπ 
Hence, the general solution of cos x = 1 is x = 2nπ, n ∈ Z.

CONCEPT:
The general solution of the equation tan x = tan α is given by: x = nπ + α, where  and n ∈ Z
Note: The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solution.
CALCULATION:

As we know that, if tan x = tan α then x = nπ + α, where  and n ∈ Z
⇒ x = nπ + (5π/6) where n ∈ Z.
So, the general solution of the given equation is given by x = nπ + (5π/6) where n ∈ Z.
Hence, the correct option is 2.

Given:
3 sin2 x - 7 sin x + 2 = 0
Concept:
Use concept of general value of sin x

Calculation:
3 sin² x - 7 sin x + 2 = 0
Put sin x = y
⇒ 3y² - 7y + 2 = 0
⇒ 3y² - 6y - y + 2 = 0
⇒ 3y(y - 2) - (y - 2) = 0
⇒ (y - 2)(3y - 1) = 0
⇒ y = 2 , 1/3
we know that range of sin x is [-1,1] then 2 is not in range than 
⇒ sin x = 1/3
⇒ x = sin^-1(1/3)
Hence the general value 
Hence the option (3) is correct.

Formula Used:
tanθ = sinθ/cosθ 
tan2θ = 2tanθ/1 - tan²θ  
Calculation:
sin θ + cos θ = x  &  cos θ - sin θ = y    
Solving equation we get, 
⇒ sinθ = (x - y)/2
⇒ cosθ = (x + y)/2
⇒ tanθ = sinθ/cosθ 
⇒ tanθ = (x - y)/(x + y)
Now, 
1 - tan²θ = 1 - (x - y)²/(x + y)²
⇒ [(x + y)² - (x - y)²]/(x + y)²
⇒ 4xy/(x + y)²

Concept:
If sin θ = sin α then θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

Calculation:
Given: 4 sin 3x = 2
⇒ sin 3x = ½
As we know that, sin (π/6) = ½
⇒ sin 3x = sin (π/6)
As we know that, if sin θ = sin α then
θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.
⇒ 3x = nπ + (- 1)ⁿ × (π/6), where n ∈ Z.
⇒ x = n × (π/3) + (- 1)ⁿ × (π/18), where n ∈ Z.

Concept:
If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

Calculation:
Given: 8 tan(2x) – 5 = 3
⇒ 8 tan(2x) = 8
⇒ tan 2x = 1
As we know that, tan (π/4) = 1
⇒ tan 2x = tan (π/4)
As we know that, if tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.
⇒ 2x = nπ + (π/4), where n ∈ Z.
⇒ x = n (π/2) + (π/8), where n ∈ Z.

Concept:
General solutions of:
cos x = cos y ⇒ x = 2nπ ± y.
sin x = sin y ⇒ x = nπ + (-1)ⁿy.
Calculation:
The given equation can be solved as:
​sin 2x + cos x = 0.
⇒ 2 sin x cos x + cos x = 0
⇒ cos x (2 sin x + 1) = 0
⇒ cos x = 0 (or) (2 sin x + 1) = 0
i.e cos x = 0 (or) sinx = -1/2
⇒ cos x = cos π/2 OR sin x = sin (-π/6)
⇒ x = 2nπ +- π/2 OR x = nπ + (-1)ⁿ × (-π/6)
Additional Information
General Solution of Trigonometric Functions

Concept:
General solution of some standard trigonometric equations:

Calculation:
Given: tan x = √3
⇒ tan x = tan (π/3)
As we know, If tan θ = tan α then θ = nπ + α
Therefore, x = nπ + (π/3)
Hence, the general solution of tan x = 

From question, the equation given is:
cos 2x + α sin x = 2α – 7
∵ [cos 2x = 1 – 2 sin² x]
⇒ 1 – 2 sin² x + α sin x – 2α + 7 = 0
⇒ -2 sin² x + α sin x – 2α + 8 = 0
⇒ 2 sin²x – αsinx + 2α – 8 = 0
⇒ (2 sin² x – 8) – α sin x + 2α = 0
⇒ 2(sin² x – 4) – α(sin x – 2) = 0
⇒ 2(sin x – 2)(sin x + 2) – α (sin x – 2) = 0
⇒(sin x – 2)[2(sin x + 2) – α] = 0
Now, (sin x – 2):
⇒ sin x – 2 = 0
∴ sin x = 2
Which is not possible.
Now, 2(sin x + 2) – α:
⇒ 2(sin x + 2) – α = 0
⇒ (2 sin x + 4) = α
⇒ 2 sin x = α – 4

Now, the range of the sin x is generally from -1 to 1
⇒ -1 < x < 1

⇒ -2 < (α – 4) < 2
∴ 2 < α < 6
Thus, the range of α is [2, 6].

tan θ = tan α
∴ θ = nπ + α