Question

Solve $sin^8\, x + cos^8 \,x = 17/32$.

• A. $\frac{n\pi}{2}$, $n \in Z$
• B. $\frac{n\pi}{2} \pm \frac{\pi}{8}$, $n \in Z$
• C. $\frac{n\pi}{2} + \frac{\pi}{4}$, $n \in Z$
• D. None of these

Answer 1

The Correct Option is B

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