Question

Solve the equation ${\sin }^{10}x+{\cos }^{10}x=\frac{29}{16}{\cos }^{4}2x.$

• A. $x=\frac{n\pi }{2}\pm \frac{\pi }{8},n\in z$
• B. $x=\frac{n\pi }{2}\pm \frac{\pi }{4},n\in z$
• C. $x=n\pi \pm \frac{\pi }{8},n\in z$
• D. $x=\frac{n\pi }{2}\pm \frac{\pi }{2},n\in z$

Answer 1

The correct option is B $x=\frac{n\pi }{2}\pm \frac{\pi }{4},n\in z$

${\sin }^{10}x+{\cos }^{10}x=\frac{29}{16}{\cos }^{4}2x\Rightarrow {\sin }^{8}x{\sin }^{2}x+{\cos }^{8}x{\cos }^{2}x=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow {\sin }^{8}x(1-{\cos }^{2}x)+{\cos }^{8}x(1-{\sin }^{2}x)=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow {\sin }^{8}x+{\cos }^{8}x-{\sin }^{8}x{\cos }^{2}x-{\cos }^{8}x{\sin }^{2}x=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow {\sin }^{8}x+{\cos }^{8}x-{\sin }^{2}x{\cos }^{2}x({\sin }^{6}x+{\cos }^{6}x)=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow {({\cos }^{4}x+{\sin }^{4}x)}^2-2{\sin }^{4}x{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x({({\sin }^{2}x+{\cos }^{2}x)}^3-3{\sin }^{2}x{\cos }^{2}x)=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow 1-{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x-{\sin }^{2}x{\cos }^{2}x(1-3{\sin }^{2}x{\cos }^{2}x)=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow 1-{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x-{\sin }^{2}x{\cos }^{2}x+3{\sin }^{4}x{\cos }^{2}x=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow 1-{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x-\frac{1}{4}{\sin }^{2}2x+3{\sin }^{4}2x(1-{\sin }^{2}x)=\frac{29}{26}{\cos }^{4}2x$

$\Rightarrow 1-\frac{5}{4}{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x+3{\sin }^{4}x-3{\sin }^{6}x=\frac{29}{16}{\cos }^{4}2x$

$\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{4}$