Question

If $si{n}^{2}4x+co{s}^{2}x=2sin4x.co{s}^{4}x$ then number of values of x satisfying, if $x\in [-2\pi ,2\pi ]$ is

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

0

$si{n}^{2}4x+co{s}^{2}x=2sin4xco{s}^{4}x\Rightarrow si{n}^{2}4x-2sin4xco{s}^{4}x+co{s}^{2}x\therefore sin4x=\frac{2co{s}^{4}x\pm \sqrt{4co{s}^{8}x-4co{s}^{2}x}}{2}sin4x=co{s}^{4}x\pm \sqrt{co{s}^{2}x(co{s}^{6}x-1)}\therefore co{s}^{6}x=1sin4x=1\Rightarrow 4x=(4n_2+1)\frac{\pi }{2}\therefore cosx=\pm 1x=(4n_2+1)\frac{\pi }{8}\therefore x=n_1\pi ,n,\in Z\therefore n_1\pi =(4n_2+1)\frac{\pi }{8}$
Not possible,
$\therefore$ No solution

The graph will help you understand better.