Question

Find the general solution of
$4{\sin }^4\theta +{\cos }^4\theta =1$

Answer 1

Given

$4{\sin }^4\theta +{\cos }^4\theta +1$

$4{\sin }^4\theta +({\cos }^2\theta +1)({\cos }^2\theta -1)=0$

$4{\sin }^4\theta -{\sin }^2\theta (1+{\cos }^2\theta )=0$

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

${\sin }^2\theta =0$

or ${\sin }^2\theta =\frac{2}{5}$

$\theta =n\pi$

${\sin }^2\theta =\frac{2}{5}$

$\sin \theta =\sqrt{\frac{2}{5}}$

$\theta ={\sin }^{-1}\sqrt{\frac{2}{5}}$

Therefore general solution of the equation is $n\pi \pm {\sin }^{-1}\left(\sqrt{\frac{2}{5}}\right)$