Question

Solve the following equations.
$\frac{3}{2}-sin2x=\sqrt{9+10sin2x}.$

Answer 1

$\frac{3}{2}-sin2x=\sqrt{9+10sin2x}$

squaring both sides we have

${(\frac{3}{2}-sin2x)}^2=(9+10sin2x)$

$\Rightarrow \frac{9}{4}+si{n}^{2}2x-3sin2x=9+10sin2x$

$\Rightarrow \frac{9}{4}-3sin2x+si{n}^{2}2x-9-sin2x=0$

$\Rightarrow -\frac{27}{4}-13sin2x+si{n}^2\left(2x\right)=0$

$\Rightarrow (sin2x-27/2)(sin2x+1/2)=0$

$\therefore sin2x-\frac{27}{2}=0$ or $sin2x=-1/2$

$\Rightarrow sin2x=\frac{27}{2}$ or $2x=2x\pi -\pi /6$

$x=x\pi -\pi /12$

(not possible) $nϵz$

$\therefore x=x\pi -\pi /12$