Question

If $3–2\cos x–4\sin x–\cos 2x+\sin 2x=0$, then $x=?(n\in Z)$

Answer 1

Given, $3–2\cos x–4\sin x–\cos 2x+\sin 2x=0$ $\Rightarrow 3–2\cos x–4\sin x–(2{\cos }^{2}x–1)+2\sin xcosx=0$
$\Rightarrow 4–4\sin x–2\cos x–2{\cos }^{2}x+2\sin x\cos x=0$
$\Rightarrow 2–2\sin x+\sin x\cos x–\cos x–{\cos }^{2}x=0$
$\Rightarrow 2(1–\sin x)–\cos x(1–\sin x)–(1–{\sin }^{2}x)=0$
$\Rightarrow (1–\sin x)[2–\cos x–1–\sin x]=0$

$\Rightarrow 1–\sin x=0\text{or}1–\cos x–\sin x=0$ $\Rightarrow \sin x=1\text{or}\cos x+\sin x=1$
$\Rightarrow \sin x=1\text{or}\cos (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$
$\Rightarrow x=n\pi +(-n{)}^n\text{or}x-\frac{\pi }{4}=2n\pi +\frac{\pi }{4}$
$\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{2}\text{or}2n\pi +\frac{\pi }{2}$
​​​​​​​Hence the correct answer is Option a.