Question

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

• A. $x=3n\pi \pm \pi$
• B. $x=2n\pi \pm \pi$
• C. $x=n\pi \pm \frac{\pi }{2}$
• D. $x=2n\pi \pm \frac{\pi }{4}$

Answer 1

Answer: (B)

$x=2n\pi \pm \pi$

$\frac{\sqrt{3}}{2}\sin x-\cos x={\cos }^{2}x$
$\Rightarrow \frac{3}{4}(1-{\cos }^{2}x)=({\cos }^{2}x+\cos x{)}^2$
$\Rightarrow 3(1-{\cos }^{2}x)-4{\cos }^{2}x(1+\cos x{)}^2=0$
$\Rightarrow (1+\cos x)[3-3\cos x-4{\cos }^{2}x-4{\cos }^{3}x]=0$
$\Rightarrow (1+\cos x)[-4{\cos }^{3}x-4{\cos }^{2}x-3\cos x+3]=0$
$\Rightarrow (1+\cos x)(\cos x-\frac{1}{2})(4{\cos }^{2}x+6\cos x+6)=0$
$\Rightarrow \cos x=-1,\cos x=\frac{1}{2},4{\cos }^{2}x+6\cos x+6=0$
$x=\pi ;x=\frac{\pi }{3};\frac{5\pi }{3};D<0$
But $\frac{5\pi }{3}$ does not satisfy the equation
$\therefore x=2n\pi +\frac{\pi }{3}$
or $x=2n\pi \pm \pi ,n\in I$