Question

The general value of $x$ for which $\cos 2x,\frac{1}{2}$ and $\sin 2$ are in $A.P.$ are given by

• A. $n\pi ,n\pi +\frac{\pi }{2}$
• B. $n\pi ,n\pi +\frac{\pi }{4}$
• C. $n\pi +\frac{\pi }{4}$
• D. $n\pi$

Answer 1

Answer: (B)

$n\pi ,n\pi +\frac{\pi }{4}$

Since $\cos 2x,\frac{1}{2},\sin 2x$ are in $A.P.$

$\Rightarrow \cos 2x+\sin 2x=2\left(\frac{1}{2}\right)=1\Rightarrow \sin 2x=1-\cos 2x=2{\sin }^{2}x\Rightarrow 2\sin x\cos x=2{\sin }^{2}x\Rightarrow \sin x(\cos x-\sin x)=0$

either $\sin x=0$ or $\cos x-\sin x=0$

$\Rightarrow x=n\pi$ or $\tan x=1$

$\Rightarrow x=n\pi$ or $x=n\pi +\frac{\pi }{4}$