Question

If $4{\sin }^{4}x+{\cos }^{4}x=1$, then
$(\text{where}x\notin n\pi ,n\in Z)$

• A. $x=n\pi \pm {\sin }^{-1}\sqrt{\frac{1}{5}},n\in Z$
• B. $x=n\pi \pm {\cos }^{-1}\sqrt{\frac{3}{5}},n\in Z$
• C. $x=n\pi \pm {\cos }^{-1}\sqrt{\frac{2}{5}},n\in Z$
• D. $x=n\pi \pm {\cos }^{-1}\sqrt{\frac{1}{5}},n\in Z$

Answer 1

The correct option is B $x=n\pi \pm {\cos }^{-1}\sqrt{\frac{3}{5}},n\in Z$

$4{\sin }^{4}x+{\cos }^{4}x=1\Rightarrow 4{\sin }^{4}x+(1-{\sin }^{2}x{)}^2=1$
Assuming ${\sin }^{2}x=t$, we get
$\Rightarrow 4t^2+(1-t{)}^2=1\Rightarrow 5t^2-2t=0\Rightarrow t(5t-2)=0\Rightarrow t=0,t=\frac{2}{5}$

When $t=0$, we get
$\Rightarrow {\sin }^{2}x=0\Rightarrow x=n\pi$
But $x\notin n\pi$

When $t=\frac{2}{5}$, we get
${\sin }^{2}x=\frac{2}{5}$

As the given options are in the form of ${\cos }^{-1}x$, so converting $\sin x$ into $\cos x.$
$\Rightarrow {\cos }^{2}x=\frac{3}{5}\therefore x=n\pi \pm {\cos }^{-1}\sqrt{\frac{3}{5}},n\in Z$