Test: Trigonometric Equations - JEE MCQ

Two results are possible as π/6 & 5π/6 comes by subtracting π/6from π, In general we have Sin∆=nπ+(-1)^n(A) here A is π/6 and by putting different values of n in integer 2 values can be obtained

sinθ = 0
⇒ sin-1 θ = 0,π,2π,...
⇒ θ = 0 + nπ → n∈Z
 The general solution for sin x = 0 will be, x = nπ, where n∈I.

The general solution of the given question is theta= 2nπ± π/6 but it is mentioned that they are lies between -π to π. So when we put n=0 we get theta =±π/6. And when we put n= 1 we get theta does not lies between -π to ÷π. So we get only two values of theta.

cos 4x = cos 2x

tanθ = tanα
θ = nπ + α
2θ = nπ + 2/θ
2(θ - 1/θ) = nπ
= θ - 1/θ = nπ/2
= θ² - (nπθ)/2 - 1 = 0
θ = nπ/2 +- [(n²π² + 4)^1/2 ]/2
θ = nπ/2 +- [(n² π² + 16)^1/2]/4

(180+theta ) lies in third quadrant...where tantheta is positive....tan(180+x)=tanx