Question

The number of solutions of the equation $1+{\sin }^{4}x={\cos }^{2}3x,x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$ is :

• A. $7$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

$1+{\sin }^{4}x={\cos }^{2}3x,x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$
As we know, range of ${\cos }^{2}3x:[0,1]$
whereas $1+{\sin }^{4}x\ge 1$
Therefore only possibilities will be
$1+{\sin }^{4}x=1={\cos }^{2}3x$
No of solution of ${\sin }^{4}x=0;x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$ is $(-2\pi ,-\pi ,0,\pi ,2\pi )\cdots \left(1\right)$
and solution of ${\cos }^{2}3x=1;3x\in [-\frac{15\pi }{2},\frac{15\pi }{2}]$ is
$3x=(-7\pi ,\cdots ,-\pi ,0,\pi ,\cdots 7\pi )\cdots \left(2\right)$
Hence, according to $\left(1\right)$ and $\left(2\right)$ there are $5$ number of solutions.