Question

${\sin }^{2}4x+{\cos }^{2}x=2\sin 4x{\cos }^{4}x$.

Solve for $x$

Answer 1

The given equation can be re-written as ${\sin }^{2}4x-2\sin 4x{\cos }^{4}x+{\cos }^{2}x=0$
Add and subtract ${\cos }^8$x
$\therefore (\sin 4x-{\cos }^{4}x{)}^2+{\cos }^{2}x(1-{\cos }^{6}x)=0$
Since both the terms are $+$ive $({\cos }^{6}x\le 1)$, above is possible only when each term is zero for the same value of x.
$\sin 4x-{\cos }^{4}x=0$ .$\left(1\right)$
and ${\cos }^{2}x(1-{\cos }^{6}x)=0$ .$\left(2\right)$
From $\left(2\right)$ $\cos x=0$ or ${\cos }^{2}x=1$
$\because z^3=1\Rightarrow z=1$ only
as other values will not be real.
Case I: If $\cos x=0$ i.e., $x=(n+\frac{1}{2})\pi$, then from $\left(1\right)$
$\sin 4(n+\frac{1}{2})\pi +0=0$
or $\sin (4n+2)\pi =0$ which is true.
$\therefore x=(n+\frac{1}{2})\pi$ $\left(3\right)$
Case II: When ${\cos }^{2}x=1$ i.e., $\sin x=0$
$\therefore x=r\pi$ then from $\left(1\right)$, $\sin 4r\pi -1=0$ or $-1=0$ which is not true. Hence the only solution is given by $\left(3\right)$.