Question

If $0\le x\le 2\pi$, then number of solutions of

Answer 1

A) ${\sin }^{2}x-\cos x=\frac{1}{4}\Rightarrow 1-{\cos }^{2}x-\cos x=\frac{1}{4}$
$\Rightarrow 4{\cos }^{2}x+4\cos x-3=0\Rightarrow (2\cos x+3)(2\cos x-1)=0$
$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-\frac{3}{2}$ (not possible)
$\Rightarrow x=\frac{\pi }{3},\frac{5\pi }{3}$
Number of solution is $2$

B) $\sin 2x=\cos 3x\Rightarrow \cos (\frac{\pi }{2}-2x)=\cos 3x$
$\Rightarrow 3x=2n\pi \pm (\frac{\pi }{2}-2x)$
$\Rightarrow x=2n\pi -\frac{\pi }{2}$ or $5x=2n\pi +\frac{\pi }{2}$
$\Rightarrow x=\frac{\pi }{10},\frac{5\pi }{10},\frac{9\pi }{10},\frac{13\pi }{10},\frac{17\pi }{10},\frac{\pi }{2},\frac{3\pi }{2}$
So number of distinct solution is $6$

C) ${\tan }^{2}x+co{t}^{2}x=2\Rightarrow {\tan }^{4}x-2{\tan }^{2}x+1=0\Rightarrow {\tan }^{2}x=1\Rightarrow \tan x=\pm 1\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$
Number of solution is $4$

D) $\sin x=\frac{1}{2},\cos x=\frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi }{6}$
Number of solution is $1$