Question

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

• A. $x=n\pi \pm \frac{\pi }{6},nϵI$
• B. $x=n\pi \pm \frac{\pi }{3},nϵI$
• C. $x=n\pi ,nϵI$
• D. $x=2n\pi +\frac{\pi }{2},nϵI$

Answer 1

Answer: (B), (C)

$x=n\pi ,nϵI$
$x=n\pi \pm \frac{\pi }{3},nϵI$

We can write the above expression as,
$2^{2{\sin }^{2}x-3\sin x+1}+\frac{2^3}{2^{2{\sin }^{2}x-3\sin x+1}}=9$
Let,
$2^{2{\sin }^{2}x-3\sin x+1}=k$
So,
$k+\frac{8}{k}=9$

$k=\frac{9\pm 7}{2}=8or1$
$\Rightarrow 2^{2{\sin }^{2}x-3\sin x+1}=8=2^3$ or $2^{2{\sin }^{2}x-3\sin x+1}=1=2^0$
$2{\sin }^{2}x-3\sin x+1=3$ or $2{\sin }^{2}x-3\sin x+1=0$
$2{\sin }^{2}x-3\sin x-2=0$ or $(2\sin x-1)(\sin x-1)=0$
$\sin x=\frac{-1}{2}or2$ or $\sin x=\frac{1}{2}Or\sin x=1$
So,
$\sin x=\frac{-1}{2},\frac{1}{2},\sin x=1$
So, $x=n\pi$ or $x=n\pi \pm \frac{\pi }{3}$