Question

Solve the following equations.
$cos\frac{3x}{2}cos\frac{x}{2}-1=\frac{1\sqrt{3}}{2}cosx.$

Answer 1

$cos\frac{3x}{2}cosx/2-1=\frac{1\sqrt{3}}{2}cosx$

$\frac{cos2x+cosx}{2}-1=\frac{\sqrt{3}}{2}cosx$ [use $cosa+cosb=cos\frac{a+b}{2}cos\frac{a-b}{2}]$

Where $a=2xb=x$

$cos2x-2=(\sqrt{3}-1)cosx$

$2co{s}^{2}x-3=(\sqrt{3}-1)cosx$

$cosx=t$

$2t^2-(\sqrt{3}-1)t-3=0$

$t=\frac{(\sqrt{3}-1)\pm \sqrt{(\sqrt{3-1}{)}^2}-\mathrm{4.2.}(-3)}{2.2}$

$t=\frac{0.7\pm 4.75}{4}$

$cosx=t=1.063$ OR $1.4125$

As cosx lies b/w $-1$ to $1$

Hence No solution