Question

Find the number of solutions of the equation $2si{n}^3+6si{n}^{2}x-sinx-3=0$ in $(0,2\pi )$.

Answer 1

Given equation $2{sin}^{3}x+6{sin}^{2}x-sinx-3=0$

Let $sinx=y,$ then the equation becomes

$2y^3+6y^2-y=3$

$(y+3)(2y^2-1)=0$

$y=-3or\frac{1}{\sqrt{2}}$

$y=sinx=-3or\frac{1}{\sqrt{2}}x={sin}^{-1}(-3)or{sin}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$x={45}^{0}or{sin}^{-1}(-3)$

hence in (0,2$\pi$) there are two values