Competition Level Test: Solution Of Triangles- 2 - JEE MCQ

The value scheme for this is shown. 

From the figure,

∴ 

2 sin x + 1 > 0  ⇒ 
The value scheme for this is shown below. 
From the figure,

∴ 

The value scheme for this is shown below. 


and un general, 

∴ 

From the graph of y = sin x, it obvious that, between 0 and 2π. 




The requried solution set is 

But for this value of x,

 which does not satisfy the given equation as it 
reduces to indeterminate form. 

From the graph of y = cos x, it is obvious that 




The required dolution is

⇒ 
⇒ 
or  
⇒ 
∴ Number of solution = 4

The given equation can be written as
 
e^{2sin x} - 4e ^sinx -1 = 0
⇒ 
⇒ sin x = In (2 +√5)
[In (2 - √5) not defined as 2-√5 is - ve]
Now, 2 + √5 > e ⇒ = In (2 + √5) > 1 = sin x > 1 
which is not possible ∴ No real solution.

We have, tan (π cos θ) = cot (π sin θ)

⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
(for n = 0 and n = -1)

⇒ 


⇒ 
or 

So that, the given equation can be written
⇒ y² - 30y + 81 = 0
⇒ 
⇒ 
⇒ 
⇒ 

⇒  

cos x. cos 2x. cos 3x = 1/4

⇒ 4cos3x • cosx • cos2x =1
⇒ 2 (cos4x + cos2x) cos 2x =1
⇒ 2(2cos² 2x - 1 +cos2x)cos2x =1 
⇒ 4cos³ 2x + 2cos² 2x - 2 cos2x-1 = 0
⇒ = 2cos² 2x(2cos 2x+ 1)- (2cos 2x+ 1) = 0 
⇒ (2cos 2x + 1) (2cos² 2x - 1) = 0
⇒ cos 4x•(2cos 2x+ 1) = 0 
⇒ It cos 4x = 0 4x = (2n₁ + 1)π/ 2
⇒ 
or 
⇒ 
⇒ 
Thus, there are six solutions.

sin⁴ x + cos⁴ x= sin x. cos x
⇒ (sin² x + cos² x)² - 2 sin² x ^. cos² x = sin x ^. cos x 
⇒ 
⇒ sin² 2x + sin 2x - 2 = 0
⇒ (sin 2x + 2)  (sin 2x - 1) = 0 ⇒ sin 2x = 1
⇒ 
⇒ 
⇒  
thus there are two solution.

Given equation is
2 sin² θ - cos 2 θ = 0
⇒ 2 sin² θ - (1 - 2sin² θ) = 0
⇒ 4 sin² θ = 1
⇒           .......(1)
Also, 2cos² θ - 3sin θ = 0

⇒ sin θ 1/2 (∵ sin θ ≠ -2)     .......(2)
From Eqs, (i) and (ii), we get sin θ = 1/2 satisfying both the given equation

Since, 2sin² x θ - 5 sin θ + 2 > 0

Since, 2 sin² x + 5 sin x - 3 = 0
⇒ (2 sin x - 1) (sin x + 3) = 0
⇒ sin x = 1/2  (∵ sin x ≠ = -3)

From figure, it is clear that the number of intersection points of y = 1/2 and y = sin x is 4. 
Hence, number of solutions of given equation in [0,3π] is 4.

Given, 4 cos² x + 6 sin² x = 5
⇒  4 (cos² x + sin² x)+2 sin² x = 5
2 sin² x = 5 - 4 

Given equation can be written as

Given  that, sin x - cos x = √2,

⇒ 
⇒ 
⇒ 
⇒ 

Since, we get know a cos x + b sin x = c has solution.,

Hence, given equation is


∴ Given equation has no solution.

Given, 3 sin²x + 10 cos x - 6 = 0
⇒ 3 (1 - cos² x) + 10 cos x - 6 = 0
⇒ 3 - 3cos² x + 9 cos x + cos x - 6 = 0
⇒ - 3 cos x (cos x - 3) + 1 (cos x - 3) = 0
⇒ (cos x - 3) (1 - 3 cos x) = 0 
⇒ cos x = 1/3
⇒ 

⇒ 

Given equation is 
sin 2x + cos 4x = 2
which is only possible,
if sin 2x = cos 4x = 1
when sin 2x = 1

For which, cos 4x = -1
∴ Given equation has no solution. 

We have, sin 3θ =  sinθ 
⇒ sin 3θ - sinθ  = 0

⇒ cos 2θ ^. sin θ = 0
⇒ cos 2θ = 0 or sin θ = 0

Thus, total number of solution = 11

Given. sin θ = sin α 
⇒  θ = nπ + (-1)ⁿ α
⇒ θ = 2nπ+α, (2n+ 1) + π - α
and cosθ = cos α
⇒ θ = 2nπ+α, 2nπ - α
∴ θ  = 2nπ+α, satisfies both equations. 

Here ‘θ’ lies in the 4th quadrant.

⇒ General value of 

Given, 3 cos 2x - 10 cos x + 7 = 0
⇒ 3 (2 cos² = x - 1) - 10 cos x + 7 = 0
⇒ 6 cos² x - 10 cos x + 4 = 0
⇒ 3 cos² x - 5 cos x + 2 = 0
⇒ (3 cos x - 2) (cos x - 1) = 0 
⇒ cos x = 1 or  cos x = 2/3,
⇒ x = 0, 2π

Hence, four values exist, when x ∈ [0, 2π]. 

sin⁶x = 1 + cos⁴x < 1
Hence, sin⁶ x = 1 + cos⁴ 3x < 1 is possible only when
sin⁶ x = 1 + cos⁴ 3x = 1




Common value of x is (2n + 1) π/2, 

The required solution

sin⁴x = 1 + tan⁸x only when
sin⁴x = 1 and 1 + tan⁸x = 1
⇒ sin²x = 1  and tan⁸x = 0
Which is never possible, since sin x and tan x vanish simultaneously.
Therefore, the given equation has no solution.