Question

Consider an equation with $x$ as a variable $7\sin 3x-2\sin 9x={\sec }^2\theta +4{\text{cosec}}^2\theta .$ Then the value of $\frac{15}{\pi }(\text{(minimum positive root)}-\text{(maximum negative root)})$ is

Answer 1

$7\sin 3x-2\sin 9x={\sec }^2\theta +4{\text{cosec}}^2\theta$
$\Rightarrow 7\sin 3x-2(3\sin 3x-4{\sin }^{3}3x)=(1+{\tan }^2\theta )+4(1+{\cot }^2\theta )$
Let $\sin 3x=y,y\in [-1,1]$
Then, $y(8y^2+1)=9+(\tan \theta -2\cot \theta {)}^2\cdots (1)$
L.H.S. will be minimum at $y=-1$
minimum value $=-9$
L.H.S. will be maximum at $y=1$
maximum value $=9$
So, equation $\left(1\right)$ has a solution only if
max of L.H.S. $=$ min of R.H.S.
$\therefore \sin 3x=1=\sin \left(\frac{\pi }{2}\right)$
$\Rightarrow x=\frac{\pi }{6}$ (minimum positive root)
Or
$\sin 3x=1=\sin \left(\frac{-3\pi }{2}\right)$
$\Rightarrow x=\frac{-\pi }{2}$ (maximum negative root)

$\text{(minimum positive root)}-\text{(maximum negative root)}$
$=\frac{\pi }{6}+\frac{\pi }{2}=\frac{2\pi }{3}$
So, $\frac{15}{\pi }(\text{(minimum positive root)}-\text{(maximum negative root)})=\frac{15}{\pi }\times \frac{2\pi }{3}=10$