I need easy solutions to these trigonometric equations:
$\sin^3x \cos x = \frac{1}{4} \text{ and }\sin^4x \cos x = \frac{1}{4}$
I need easy solutions to these trigonometric equations:
$\sin^3x \cos x = \frac{1}{4} \text{ and }\sin^4x \cos x = \frac{1}{4}$
The first equation can be written as $-\frac{\sin(4x)}{8} + \frac{\sin(2x)}{4} = \frac{1}{4}$ Note that if $\sin(2x)=1$, $\sin(4x)=0$.
Alternatively, write $\sin(x) = (z-1/z)/(2 i)$ and $\cos(x) = (z+1/z)/2$ and factor.
As far as I can tell, the second equation has no "easy" solution.