Question

Solve the following equations.
$4cosx-2cos2x-cos4x=1.$

Answer 1

$4cosx-2cos2x-cos4x=1$

$\Rightarrow 4cosx-cos2x-cos4x=1+cos2x$

$\Rightarrow 4cosx-2cos3xcosx=2co{s}^{2}x$

Say $cosx=t$

$\Rightarrow 4t-2t(4t^3-3t)=2t^2$

$\Rightarrow 4t-8t^4+6t^2=2t^2$

$\Rightarrow 8t^4-4t^2-4t=0$

$\Rightarrow 2t^4-t^2-t=0$

$\Rightarrow t(t-1)(2t^2+2t+1)=0$

$\therefore t=0$ or $t=1$

$\Rightarrow cosx=0$ or $\Rightarrow cosx=1$

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

$nϵ$ integer $mϵ$ integer