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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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No, if $\sin x = 1$, $\cos x = 0$. The two equations are separate. – Robert Israel Mar 7 '12 at 19:27
if sin(x)=1, cos(x)=1/4 – dato datuashvili Mar 7 '12 at 19:36
but at the same time there is no such x which satisfy both – dato datuashvili Mar 7 '12 at 19:36
Assuming these are two separate problems: For the second, use cos^2+sin^2=1 to write everything in terms of cos, then make a substitution u=cos x so you get a polynomial in u. Find the roots of it and see which x's give you those u's. If you're trying to solve them simultaneously use the finitely many solutions of the second and plug into the first and see if any of them work. – user16544 Mar 7 '12 at 19:37
in this case he will get polynomial with higher degree is not it? – dato datuashvili Mar 7 '12 at 19:39
up vote 5 down vote accepted

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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The second one requires the solution of a quintinc: $$\alpha^5-2\alpha^3+\alpha-1/4=0$$. Then $\alpha = \cos x$ – Pedro Tamaroff Mar 7 '12 at 19:49
They are separate. – Adam Mar 7 '12 at 20:08
Dear Robert, How can I prove that there are no other solutions except sin(2x)=1? – Adam Mar 11 '12 at 23:56
I didn't say there are no other solutions. There are. But they are not "easy" solutions. The other solutions of the first equation have $\sin(2x) = s$ where $s^3+s^2+s-1=0$. This cubic has one real root, approximately $0.5436890127$. – Robert Israel Mar 12 '12 at 1:09
Show that the derivative of the left side is nonnegative on that interval. – Robert Israel Mar 13 '12 at 6:15

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