JEE Advanced Level Test: Trigonometric Equations- 1 - Airforce X Y / Indian Navy SSR MCQ

2⋅cos2x=2(2cos²x−1) x+cos2x
=2cos² x − 4 
x=2cos²x (cosx=±1)/x
=nπ, n∈I

4sinqcosq - 2cosq - 2√3sinq + √3 = 0
2cosq(2sinq - 1) - √3(2sinq - 1) = 0
(2sinq - 1) (2cosq - √3) = 0
sinq = 1/2 , cosq = √3/2
q = π/6, 5π/6. , q = π/6, 11π/6
q = {π/6 , 5π/6, 11π/6}

sin x tan 4x = cos x

sin x sin 4x = cos x cos 4x

cos x cos 4x - sin x sin 4x = 0

cos(x + 4x) = 0

cos 5x = 0

5x = pi/2 or 5x = 3pi/2 or 5x = 5pi/2

or 5x = 7pi/2 or 5x = 9pi/2 

x = pi/10 or x = 3pi/10 or x = pi/2 

or x = 7pi/10 or x = 9pi/10.

number of solutions is 5.

2 sin θ + tan θ = 0⇒2 sin θ + sin θcos θ = 0⇒2 sin θ . cos θ + sin θ = 0⇒sin θ[2 cos θ + 1] = 0⇒sin θ = 0  or  2 cos θ + 1 = 0⇒sin θ = 0  or cos θ = −12⇒sin θ = 0  or cos θ = −cos (π/3)⇒sin θ = 0  or cos θ = cos (2π/3)⇒θ = nπ or θ = 2mπ±2π3

We have tan θ = -1 and cos θ =1/√2 .
So, θ lies in IV quadrant.
θ = 7/4
So, general solution is θ = 7π/4 + 2 n π, n∈ Z

2cos²(π+x)+3sin(π+x)
=2cos²(x)−3sin(x)
=(1−sin²(x))⋅2−3sin(x)
∴sin(x)=−2,sin(x)=1/2
sin(x)=−2:None
and sin(x) = 1/2: x=π/6+2πn, x=5π/6+2πn
∴x=π/6 or 5π/6

We have, sin7x+sinx+sin4x=0
2sin4xcos3x+sin4x=0
⇒sin4x(2cos3x+1)=0
⇒sin4x=0,cos3x=−1/2
⇒4x=2kπ
⇒2kπ/4
​=kπ/2
if k=0⇒x=0
if k=1⇒x=π/2
cos3x=−1/2
⇒ Again 3x=±2π/3, 5π/18
3x=2π/3=2kπ
⇒x=2π/9+2kπ/3
k=0, x=2π/9
k=1, x=2π/9+2π/3=8π/9
out of range
x=(−2π)/3 + 2kπ (out of range)
∴ For x=[0,π/2] we get 5
solution:
{0,2π/9,π/2} {2π/3,5π/18}
 

cot3θ−cotθ=0 
(3cotθ−cot³ θ)/(1−3cot² θ)=cotθ 
cotθ(3−cot² θ)=cotθ(1−3cot² θ) 
⟹ cotθ=0 
⟹ cotθ=cotπ/2
⟹ θ=(2n−1)π/2

tan(A-B)=(tanA-tanB)/(1+tanA*tanB) THEN
tan(3x-2x)=( tan3x−tan2x)/(1+tan3xtan2x)
tanx = 1
x =π/4
x =nπ +π/4

we can write sin²x=[(1−cos2x)]/2
(a1)+(a3)/2=0
(a2)−(a3)/2=0
These two equations can have infinite roots.

Dear Sanchit

4 cosec ² (π (a+x) ) + a² - 4a = 0

4 cosec ² (π (a+x) ) =- a² + 4a 

L .H .S is always greater than 4 so for real solution

 -a² + 4a ≥4

or (a-2)²≤0

so only a=2 is possible

Squaring both sides

cos^2 x=cos^2 x + 4sin^2x-4sinxcosx

4sinxcox=4sin^2 x

sin x=cos x

Principal solution = pi/4=45 degrees

General Solution = 2nπ+π/4