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I need easy solutions to these trigonometric equations:

$\sin^3x \cos x = \frac{1}{4} \text{ and }\sin^4x \cos x = \frac{1}{4}$

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    I assumed this was what was intended and the roots would jump out but they're not not obvious you are right. Using sage two of the roots are real with absolute value <= 1 so it does have solutions at least.2012-03-07

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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.

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    Thanks. How did you get the cubic equation?2012-03-13