Question

Solve the following equations.
$si{n}^2\left(1.5x\right)+si{n}^2(\frac{\pi }{4}-2.5x)=si{n}^2\left(5.5x\right)+si{n}^2(\frac{\pi }{4}-6.5x).$

Answer 1

$si{n}^2\left(1.5x\right)+si{n}^2(\frac{\pi }{4}-2.5x)=si{n}^2\left(5.5x\right)+si{n}^2(\frac{\pi }{4}-6.5x)$

$si{n}^2\left(1.5x\right)-si{n}^2\left(5.5x\right)=si{n}^2(\frac{\pi }{4}-6.5x)-si{n}^2\left(\frac{\pi }{4}2.5x\right)$

$si{n}^{2}A-si{n}^{2}B=sin(A-B)sin(A+B)$

$\Rightarrow sin(-4x)sin\left(7x\right)=sin(-4x)sin(\frac{\pi }{2}-9x)$

$sin4x=0$ is a solution

$4x=n\pi \Rightarrow x=\frac{n\pi }{4},n=0,\pm 1,\pm 2...$

$sin7x=sin(\frac{\pi }{2}-9x)$

If sin A = sin B

$\Rightarrow A=B$ or $A+B={180}^{\circ }$

$A=B=7x=\frac{\pi }{2}-9x\Rightarrow \frac{\pi }{2}=16x\Rightarrow x=\frac{\pi }{32}$

$:A+B=\pi :\Rightarrow 7x+\frac{\pi }{2}-9y=\pi$

$\Rightarrow -2x=\frac{-\pi }{2}\Rightarrow x=\frac{-\pi }{4}$