Question

Which of the following is not the general solution of $2^{\cos 2x}+1=3\times 2^{-{\sin }^{2}x}$ ?

• A. $(n-\frac{1}{2})\pi ,n\in Z$
• B. $(n+\frac{1}{2})\pi ,n\in Z$
• C. $n\pi ,n\in Z$
• D. None of these

Answer 1

The correct option is D None of these

$2^{\cos 2x}+1=3\times 2^{-{\sin }^{2}x}\cdots \left(1\right)$

$2^{\cos 2x}=2^{1-2{\sin }^{2}x}=2\times (2^{-{\sin }^{2}x}{)}^2$
Put $2^{-{\sin }^{2}x}=t$
Therefore, Eq. $\left(1\right)$ becomes
$2t^2-3t+1=0$
$\Rightarrow t=1,\frac{1}{2}$

$2^{-{\sin }^{2}x}=1=2^0$
$\Rightarrow {\sin }^{2}x=0$
$\Rightarrow x=n\pi ,n\in Z$

$2^{-{\sin }^{2}x}=\frac{1}{2}=2^{-1}$
$\Rightarrow {\sin }^{2}x=1$
$\Rightarrow x=n\pi \pm \frac{\pi }{2},n\in Z$