Question

If $1+{\sin }^{4}x={\cos }^{2}3x$, when $xϵ[-\frac{5\pi }{2},\frac{5\pi }{2}]$, then the total number of solutions are,

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is D 5

$1+{\sin }^{4}x={\cos }^{2}3x$,

$1-{\cos }^{2}3x=-{\sin }^{4}x$

$si{n}^{2}3x+si{n}^{4}x=0$

$\Rightarrow \left(si{n}^{2}x\right)(3-4si{n}^{2}x{)}^2+si{n}^{4}x=0$

$\Rightarrow si{n}^{2}x(9-24si{n}^{2}x+16si{n}^{4}x+si{n}^{2}x)=0$

$\Rightarrow si{n}^{2}x(9-23si{n}^{2}x+16si{n}^{4}x)=0$

The second factor is a quadratic in $si{n}^{2}x$ and has imaginary roots.

Hence,
$sinx=0$

Hence, $x=-2\pi ,-\pi ,0,\pi ,2\pi$

$\therefore$ ${}^{\prime }{x}^{\prime }$ has a total of $5$ solutions

Hence option ${}^{\prime }{D}^{\prime }$ is the answer.