Question

Solve the following equation:
$4{\sin }^{4}x+12{\cos }^{2}x=7$

Answer 1

${4\sin }^{2}x+12(1-{\sin }^{2}x)=7$

${4\sin }^{2}x-{12\sin }^{2}x+5=0$

${4\sin }^{2}x-{10\sin }^{2}x-{2\sin }^{2}x+5=0$

$({2\sin }^{2}x-1)({2\sin }^{2}x-5)=0$

So, $({2\sin }^{2}x-5)=0$ is not possible.

$\therefore$ $({2\sin }^{2}x-1)=0,\sin x=\pm \frac{1}{\sqrt{2}}$

So, $x=n\pi +{(-1)}^n\left(\frac{\pi }{4}\right)$