Question

If $si{n}^{10}x+co{s}^{10}x=\frac{29}{16}co{s}^{4}2x.$ then x=

• A. $x=\frac{n\pi }{4}$
• B. $x=\frac{n\pi }{4}+\frac{\pi }{8}$
• C. $x=\frac{n\pi }{4}+\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is D None of these

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

$({\sin }^{2}x{)}^5+({\cos }^{2}x{)}^5=\frac{29}{16}{\cos }^{4}2x$

${\left(\frac{1-\cos 2x}{2}\right)}^5+{\left(\frac{1+\cos 2x}{2}\right)}^5=\frac{29}{16}{\cos }^{4}2x$

$\cos 2x=t$

${\left(\frac{1-t}{2}\right)}^5+{\left(\frac{1+t}{2}\right)}^5=\frac{29}{16}{t}^4$

$\frac{(1-t{)}^5}{32}+\frac{(1+t{)}^5}{32}=\frac{29}{16}{t}^4$

$(1-t{)}^5+(1+t{)}^5=58t^4$

$24t^4-10t^2-1=0$

Factors

$(12t^2+1)(2t^2-1)=0$

$12t^2+1=0$

No real values of t

$2t-1=0$

$t^2=\frac{1}{2}$

${\cos }^{2}2x=\frac{1}{2}$

$2{\cos }^{2}2x=1$

$1+\cos 4x=1$

$\cos 4x=0$

$4x=(2n+1)\frac{\pi }{2}$

$x=(2n+1)\frac{\pi }{8}$

$n\in$ integer

$x=\frac{(2n+1)\times \pi }{8}$

$=\frac{2n\pi +\pi }{8}-\frac{n\pi }{4}+\frac{\pi }{4}$

Option D. None of these.