Question

Solve the equation $\frac{\sqrt{3}}{2}\sin x-\cos x={\cos }^{2}x$

• A. $x=(n+1)\pi$ $nϵZ$
• B. $x=2n\pi \pm \frac{\pi }{6}$, $nϵZ$
• C. $x=(2n+1)\pi$ $nϵZ$
• D. $x=2n\pi \pm \frac{\pi }{3}$, $nϵZ$

Answer 1

Answer: (C), (D)

$x=(2n+1)\pi$ $nϵZ$
$x=2n\pi \pm \frac{\pi }{3}$, $nϵZ$

We have $\sqrt{3}\sin x=2(\cos x+{\cos }^{2}x)$

Squaring both sides, we get
$3{\sin }^{2}x=4({\cos }^{2}x+{\cos }^{4}x+2{\cos }^{3}x)$

$3(1-{\cos }^{2}x)=4{\cos }^{2}x+4{\cos }^{4}x+8{\cos }^{3}x$

$4{\cos }^{4}x+8{\cos }^{3}x+7{\cos }^{2}x-3=0$

$(\cos x+1)(2\cos x-1)(2{\cos }^{2}x+3\cos x+3)=0$

For $\cos x=-1$, $x=(2n+1)\pi :n\in Z$

For $\cos x=\frac{1}{2}$, $x=2n\pi \pm \frac{\pi }{3}:n\in Z$