Question

Which of the following set of values of $x$ satisfies the equation $2^{2{\sin }^{2}x-3\sin x+1}+2^{2-2{\sin }^{2}x+3\sin x}=9$
(where $n\in Z$)

• A. $n\pi +\frac{\pi }{4}$
• B. $n\pi +\frac{\pi }{6}$
• C. $n\pi -\frac{\pi }{6}$
• D. $2n\pi +\frac{\pi }{2}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $n\pi +\frac{\pi }{6}$
C $n\pi -\frac{\pi }{6}$
D $2n\pi +\frac{\pi }{2}$

$2^{2{\sin }^{2}x-3\sin x+1}+2^{2-2{\sin }^{2}x+3\sin x}=9$
$2^{2{\sin }^{2}x-3\sin x+1}+2^{3-(2{\sin }^{2}x-3\sin x+1)}=9$
Let $2^{2{\sin }^{2}x-3\sin x+1}=t$
$\therefore t+\frac{8}{t}=9\Rightarrow t^2-9t+8=0\Rightarrow t=1,8$

When $t=1$
$2^{2{\sin }^{2}x-3\sin x+1}=1$
$\Rightarrow 2{\sin }^{2}x-3\sin x+1=0\Rightarrow (2\sin x-1)(\sin x-1)=0\Rightarrow \sin x=\frac{1}{2},1$

When $t=8$
$2^{2{\sin }^{2}x-3\sin x+1}=8\Rightarrow 2{\sin }^{2}x-3\sin x+1=3\Rightarrow 2{\sin }^{2}x-3\sin x-2=0\Rightarrow (2\sin x+1)(\sin x-2)=0\Rightarrow \sin x=-\frac{1}{2}(\because \sin x\in [-1,1\left]\right)$
From both, we get
$\sin x=\pm \frac{1}{2},1\Rightarrow x=n\pi \pm \frac{\pi }{6},x=2n\pi +\frac{\pi }{2},n\in Z$