Question

Solve the following equation:
${\cos }^{2}2x+{\cos }^{2}3x=1$

Answer 1

${\cos }^{2}2x+{\cos }^{2}3x=1$

${({2\cos }^{2}x-1)}^2+{({4\cos }^{3}x-3\cos x)}^2=1$

$1-{4\cos }^{4}x-{4\cos }^{4}x+{16\cos }^{6}x-{24\cos }^{4}x=1$

${16\cos }^{6}x-{20\cos }^{4}x+{5\cos }^{3}x=0$

${16\cos }^{4}x-{20\cos }^{2}x+{5\cos }^{3}x=0$

Now,

${\cos }^{2}x=\frac{20\pm \sqrt{{\left(20\right)}^2-4\times 16\times 5}}{32}=\frac{5\pm \sqrt{5}}{8}$

Let $a=\frac{5+\sqrt{5}}{8}$ and $b=\frac{5-\sqrt{5}}{8}.$

So, $\cos x=\pm \sqrt{a},$ $\cos x=\pm \sqrt{b}$ and $\cos x=0.$

So, $x=2n\pi \pm {cos}^{-1}\sqrt{a},x=2n\pi \pm {cos}^{-1}\sqrt{b}$ and $x=(2n+1)\frac{\pi }{2}.$