Question

Solve the following equation:
${\sin }^{6}x+{\cos }^{6}x=\frac{7}{16}.$

Answer 1

${\sin }^{6}x+{(1-{\sin }^{2}x)}^3=\frac{7}{16}$

Let $\sin x=z.$

$z^6+1-z^6-{3z}^2(1-z^2)=\frac{7}{16}$

$16-16({3z}^2+{3z}^4)=7$

${48z}^4+{48z}^2-9=9$

$z^2=\frac{-48\pm \sqrt{4032}}{96}=\frac{-6\pm \sqrt{63}}{12}.$

So, $z^2=\frac{-6\pm \sqrt{63}}{12}=a$ $(let.)$

$\therefore$ So, $x=n\pi \pm {(-1)}^n{\sin }^{-1}\left(\sqrt{a}\right).$