Question

# The solution of $$\displaystyle \sin ^{ 8 }{ x } +\cos ^{ 8 }{ x } =\frac { 17 }{ 32 }$$ is

A
nπ2±π8
B
nπ±π4
C
nπ±π8
D
No solution

Solution

## The correct option is A $$\displaystyle \frac { n\pi }{ 2 } \pm \frac { \pi }{ 8 }$$$$\sin ^{8}(x)+\cos ^{8}(x)$$ $$=(\sin ^{4}x+\cos ^{4}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=((\sin ^{2}x+\cos ^{2}x)^{2}-2\sin ^{2}x\cos ^{2}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=(1-\dfrac{\sin ^{2}2x}{2})^{2}-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{4}-\sin ^{2}2x-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{8}-\sin ^{2}(2x)$$ $$=\dfrac{17}{32}$$ Therefore  Let $$\sin ^{2}(2x)=t$$ Hence $$\dfrac{t^{2}}{8}-t+\dfrac{15}{32}=0$$   $$4t^{2}-32t+15=0$$ $$\Rightarrow t=\dfrac{32\pm\sqrt{1024-240}}{8}$$   $$\Rightarrow t=\dfrac{32\pm4\sqrt{64-15}}{8}$$   $$\Rightarrow t=\dfrac{8\pm\sqrt{49}}{2}$$   $$\Rightarrow t=\dfrac{15}{2}$$ or $$t=\dfrac{1}{2}$$   Now, $$t=\dfrac{15}{2}$$ is not possible.   $$\sin ^{2}(2x)\epsilon[0,1]$$ Therefore  $$\sin ^{2}2x=\dfrac{1}{2}$$ $$\sin (2x)=\pm\dfrac{1}{\sqrt{2}}$$   $$2x=\dfrac{(2n+1)\pi}{4}$$ Or  $$2x=\dfrac{n\pi}{2}\pm\dfrac{\pi}{8}$$ $$\Rightarrow x=\dfrac{n\pi}{4}\pm\dfrac{\pi}{8}$$Maths

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