Question

Solve $2{\sin }^{3}x=\cos x$

Answer 1

The given equation can be written as $2{\sin }^3\left(x\right)=\sqrt{1-{\sin }^2\left(x\right)}$
$4{\sin }^6\left(x\right)=1-{\sin }^2\left(x\right)$
$4{\sin }^{6}x+{\sin }^2\left(x\right)-1=0$
Let ${\sin }^2\left(x\right)=t$
The the equation reduces to
$4t^3+t-1=0$
$(t-\frac{1}{2})(4t^2+2t+2)=0$
$(2t-1)(2t^2+t+1)=0$
$t=\frac{1}{2}$ or $2t^2+t+1=0$
Now considering
$2t^2+t+1=0$
$b^2-4ac=-7$
Hence $D<0$.
Therefore no real roots.
Hence we are left with $t=\frac{1}{2}$
${\sin }^2=\frac{1}{2}$
$\sin \left(x\right)=\pm \frac{1}{\sqrt{2}}$
$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{-\pi }{4}$
Now
$x=\frac{-\pi }{4},\frac{3\pi }{4}$ does not satisfy the equation.
Hence there are only two roots to the above equation.
They are $x=\frac{\pi }{4},\frac{5\pi }{4}$.