Question

Solve the following equations.
$si{n}^2(\frac{\pi }{8}+x)=sinx+si{n}^2(\frac{\pi }{8}-x).$

Answer 1

$si{n}^2(\pi /8+x)=sinx+si{n}^2(\pi /8-x)$

$si{n}^2(\pi /8+x)-si{n}^2(\pi /8-x)=sinx$

Apply $si{n}^2(a+b)-si{n}^2(a-b)=sin2asin2b$

$sin\left[2\left(\pi /8\right)\right]sin2x=sinx$

$\frac{2sinxcosx}{\sqrt{2}}=sinx$

$sin2x=2sinxcosx$

$sin\pi /4=1/\sqrt{2}$

$sinx(cosx-\frac{1}{\sqrt{2}})=0$

$sinx=0\Rightarrow x=x\pi xϵ$ Integer

OR $cosx=\frac{1}{\sqrt{2}}$

$x=\pi /4,-\pi /4,7\pi /4,...$

In $1^{st}\times 4^{th}$ quadrant