Question

The general solution of $\sin 2x=4\cos x$ is

• A. $(2n+1)\frac{\pi }{2},\forall n\in Z$
• B. $0\pi ,\forall n\in Z$
• C. No Solution
• D. $2n\pi \forall n\in Z$

Answer 1

Answer: (A)

$(2n+1)\frac{\pi }{2},\forall n\in Z$

$\sin 2x=4\cos x$

$\Rightarrow 2\sin x\cos x=4\cos x$

$\Rightarrow \sin x\cos x=2\cos x$

$\Rightarrow \sin x\cos x-2\cos x=0$

$\Rightarrow \cos x(\sin x-2)=0$

Now $\sin x-2\ne 0$

So $\cos x=0$

hence $x=(2n+1)\frac{\pi }{2}$