Question

Solve the following equations.
$cos2x-3cosx=4co{s}^2\frac{x}{2}.$

Answer 1

$cos2x-3cosx=4co{s}^{2}x/2$

$\Rightarrow cos2x-3cosx=2(1+cosx)$

$\Rightarrow cos2x-3cosx=2+2cosx$

$\Rightarrow 2co{s}^{2}x-1-3cosx-2-2cosx=0$

$\Rightarrow 2co{s}^{2}x-5cosx-3=0$

$\Rightarrow 2co{s}^{2}x-cosx+cosx-3=0$

$\Rightarrow 2cosx(cosx-3)+(cosx-3)=0$

$cosx=-1/2$ or $cosx=3$

$\Rightarrow x=2m\pi \pm 2\pi /3$

cos x cannot take values greater than for less then $-1$

$mϵ$ integer