Question

If $\frac{1}{6}$sinθ.cosθ.tanθ are in G.P, then the solution set θ is

• A. $2n\pi\pm\frac{\pi}{6}$
• B. $2n\pi\pm\frac{\pi}{3}$
• C. $n\pi+(-1)^n\frac{\pi}{3}$
• D. $n\pi+\frac{\pi}{3}$

Answer 1

Answer: (B)

$2n\pi\pm\frac{\pi}{3}$

Answer 2

The correct answer is option (B): $2n\pi\pm\frac{\pi}{3}$

According to the given condition, if a,b and c are in GP,we have b2=ac

Similarly,

$tan\theta sin\theta\times\frac{1}{6}=cos^2\theta$

$\Rightarrow sin^2\theta=6cos^3\theta$

$\Rightarrow 6cos^3\theta+cos^2\theta-1=0$

Since $cos\theta=\frac{1}{2}$​, the general solution becomes $2n\pi\pm\frac{\pi}{3}$