Question

Find all values between $0$ and $\pi$ which satisfies the equation : $si{n}^{8}c+co{s}^{8}x=\frac{17}{16}co{s}^{2}2x$

• A. $x=\frac{n\pi }{2}\pm (-1{)}^n\frac{\pi }{8};nϵz$
• B. $x=\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{8};nϵz$
• C. $x=\frac{n\pi }{2}-(-1{)}^n\frac{\pi }{8};nϵz$
• D. $x=\frac{n\pi }{4}\pm (-1{)}^n\frac{\pi }{8};nϵz$

Answer 1

The correct option is A $x=\frac{n\pi }{2}\pm (-1{)}^n\frac{\pi }{8};nϵz$

${\sin }^{8}x+{\cos }^{8}x=\frac{17}{32}$
$\Rightarrow {({\sin }^{4}x+{\cos }^{4}x)}^2-2{\sin }^{4}x{\cos }^{4}x=\frac{17}{32}$
$\Rightarrow {(1-2{\sin }^{2}x{\cos }^{2}x)}^2-2{\sin }^{4}x{\cos }^{4}x=\frac{17}{32}$
$\Rightarrow 1-4{\sin }^{2}x{\cos }^{2}x+2{\sin }^{4}x{\cos }^{4}x=\frac{17}{32}$
$\Rightarrow {1-\left(2\sin x\cos x\right)}^2+\frac{1}{8}{\left(2\sin x\cos x\right)}^4-\frac{17}{32}$
$\Rightarrow 1-{\sin }^{2}x+\frac{1}{8}{\sin }^{4}2x-\frac{17}{32}=0\Rightarrow 4{\sin }^{4}2x-32{\sin }^{2}2x+15=0$
$\Rightarrow 2{\sin }^{2}2x-1=0[\because {\sin }^{2}2x\ne \frac{15}{2}]$
$\Rightarrow {\sin }^{2}2x=\frac{1}{2}\Rightarrow 2x=n\pi \pm \frac{\pi }{4}\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{8}$