Question

If $\sin 2x=4\cos x$, then $x$ is equal to

• A. $\left(\frac{n\mathrm{\pi }}{2}\right)\pm \frac{\mathrm{\pi }}{4};n\in Z$
• B. no value
• C. $n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{2};\mathrm{n}\in \mathrm{Z}$
• D. $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2};\mathrm{n}\in \mathrm{Z}$

Answer 1

The correct option is D

$2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2};\mathrm{n}\in \mathrm{Z}$

Explanation for the correct option.

Given that,$\sin 2x=4\cos x$
$\begin{array}{rcl}2\sin x\cos x-4\cos x& =& 0\\ 2\cos x(\sin x-2)& =& 0\\ 2\cos x& =& 0,\sin x-2=0\\ \cos x& =& 0,\sin x=2\end{array}$
Since $-1\le \sin \theta \le 1$

So, $\sin \theta =2$ not possible.
Therefore, $\cos x=0$
$\Rightarrow x=2n\pi \pm \frac{\mathrm{\pi }}{2},n\in Z$

Hence, option C is correct.