Question

Solve the following equations.
$4co{s}^{2}6x+16co{s}^{2}3x=13.$

Answer 1

$4co{s}^{2}6x+16co{s}^{2}3x=13$

$\Rightarrow 4(2co{s}^{2}3x-1{)}^2+16co{s}^{2}3x=13$

Let $co{s}^{2}3x=t$

$\Rightarrow 4(2t-1{)}^2+16t=13$

$\Rightarrow 4(4t^2+1-4t)+16t=13$

$\Rightarrow 16t^2+4-16t+16t=13$

$\Rightarrow 16t^2=9$

$t^2=9/16$

$t=\pm 3/4$

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

$co{s}^{2}3x=\pm \sqrt{3}/2$

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

$\Rightarrow 3x=2x\pi \pm \pi /6$

$\Rightarrow x=\frac{3x\pi }{3}\pm \pi /18$