Question

The number of real solutions of the equation ${\left(3+\sin x\right)}^2+2{\cos }^{16}x=4;x\in \left[0,6\pi \right],$ is

Answer 1

$(3+\sin x{)}^2+2{\cos }^{16}x=4$

$(3+\sin x{)}^2>0$

$-1\le \sin x\le 1$

$2\le 3+\sin x\le 4$

From 1,2 and 3

cos x has to be zero and $(3+\sin x)$ has to be minimum for the given condition to satisfy

$\to 3+\sin x=2$ and $\cos =0$

$\sin x=-1$ and $\cos x=0$

$x=\left(\frac{x\pi }{2}\right)$------4

$\sin x=-1$
$\to x\in (4x+3)\frac{\pi }{2}$-------5

from 4 and 5

$x\in (4x+3)\frac{\pi }{2}$

$\therefore$ if $x\in [0,6\pi ]$

$x$ can take the value

$(\frac{3\pi }{2},\frac{7\pi }{2},\frac{11\pi }{2})$

$\therefore$ there are $3$ real solution