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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.2012-03-07
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    if sin(x)=1, cos(x)=1/42012-03-07
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    but at the same time there is no such x which satisfy both2012-03-07
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    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.2012-03-07
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    in this case he will get polynomial with higher degree is not it?2012-03-07
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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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    The second one requires the solution of a quintinc: $$\alpha^5-2\alpha^3+\alpha-1/4=0$$. Then $\alpha = \cos x$2012-03-07
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    They are separate.2012-03-07
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    Dear Robert, How can I prove that there are no other solutions except sin(2x)=1?2012-03-11
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    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 $0.5436890127$.2012-03-12
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    Thank you. How can I prove that it has no other solution in [0;Pi/4]?2012-03-12
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    Show that the derivative of the left side is nonnegative on that interval.2012-03-13
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    Thanks. How did you get the cubic equation?2012-03-13