Question

Solve the following equations.
$co{s}^{6}x-si{n}^{6}x=\frac{13}{8}co{s}^{2}2x.$

Answer 1

$co{s}^{6}x-si{n}^{6}x=\frac{13}{8}co{s}^{2}2x$

$\Rightarrow (co{s}^{3}x{)}^2-(si{n}^{3}x{)}^2=\left(13/8\right)co{s}^{2}2x$

$\Rightarrow (co{s}^{3}x-si{n}^{3}x)(co{s}^{3}x+si{n}^{3}x)=13/8co{s}^{2}2x$

$\Rightarrow (cosx-sinx)(cosx+sinx)(1-sin.cosx)(1+sinx.cosx)=13/8co{s}^{2}2x$

$\Rightarrow (co{s}^{2}x-si{n}^{2}x)(1-si{n}^{2}x.co{s}^{2}x)=13/8co{s}^{2}x$

$\Rightarrow cos2x(1-\frac{si{n}^{2}2x}{4})=13/8co{s}^{2}2x$

$\Rightarrow cos2x\frac{(4-si{n}^{2}2x)}{4}=\frac{13}{8}co{s}^{2}2x$

say

$t=cos2x$

$\Rightarrow 2t(4-1+t^2)=13t^2$

$\Rightarrow 2t^3-13t^2+6t=0$

$\Rightarrow t(2t-1)(t-6)=0$

$\therefore t=0$

$\Rightarrow cos2x=0$

$\Rightarrow x=(2x+1)\pi /4$

$t=1/2$

$\Rightarrow x=x\pi \pm \frac{\pi }{6}$

$t=1/6$

$\Rightarrow cos2x=1/6$

$\Rightarrow 2x=2x\pi \pm co{s}^{-1}1/6$

$\Rightarrow x=x\pi \pm \frac{co{s}^{-1}1/6}{2}$