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I need easy solutions to these trigonometric equations:

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

Martin Sleziak
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Adam
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  • No, if $\sin x = 1$, $\cos x = 0$. The two equations are separate. – Robert Israel Mar 07 '12 at 19:27
  • if sin(x)=1, cos(x)=1/4 – dato datuashvili Mar 07 '12 at 19:36
  • but at the same time there is no such x which satisfy both – dato datuashvili Mar 07 '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. –  Mar 07 '12 at 19:37
  • in this case he will get polynomial with higher degree is not it? – dato datuashvili Mar 07 '12 at 19:39
  • 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. –  Mar 07 '12 at 19:50

1 Answers1

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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.

Robert Israel
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