Question

Number of values of ${}^{\prime }{x}^{\prime }\in (-2\pi ,2\pi )$ satisfying the equation $2^{{\sin }^{2}x}+4\cdot 2^{{\cos }^{2}x}=6$ is

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is C $4$

$2^{si{n}^{2}x}+2^{2+co{s}^{2}x}=6$
$2^{si{n}^{2}x}+{8.2}^{-si{n}^2\left(x\right)}=6$
Let
$2^{-si{n}^2\left(x\right)}=t$.
Hence
$t+8.\frac{1}{t}=6$.
Hence
$t^2+8=6t$
Or
$t^2-6t+8=0$
Or
$(t-4)(t-2)=0$
Or
$t=4$ and $t=2$.
Hence
$2^{si{n}^2\left(x\right)}=2^2$ ... not possible.
Hence
$2^{si{n}^2\left(x\right)}=2^1$
Or
$si{n}^2\left(x\right)=1$
Or
$x=\frac{(2n-1)\pi }{2}$. where $nϵI$.
Hence we get 4 solutions in $[-2\pi ,2\pi ]$.