Question

If $1+{\sin }^{4}x={\cos }^{2}3x$, when $xϵ[-\frac{5\pi }{2},\frac{5\pi }{2}]$, then the least solution is,

• A. $-\frac{\pi }{2}$
• B. $-\pi$
• C. $-\frac{3\pi }{2}$
• D. $-2\pi$

Answer 1

The correct option is D $-2\pi$

${\sin }^{2}3x+{\sin }^{4}x=0$
$\Rightarrow \left({\sin }^{2}x\right)(3-4{\sin }^{2}x{)}^2+{\sin }^{4}x=0$
$\Rightarrow {\sin }^{2}x(9-24{\sin }^{2}x+16{\sin }^{4}x+{\sin }^{2}x)=0$
$\Rightarrow {\sin }^{2}x(9-23{\sin }^{2}x+16{\sin }^{4}x)=0$
The second factor is a quadratic in ${\sin }^{2}x$ and has imaginary roots.
Hence,
$\sin x=0$
Hence, $x=-2\pi ,-\pi ,0,\pi ,2\pi$
Hence, smallest value of $x$ is $-2\pi$