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The question is : Find the general solution of this equation $ \sin (5\phi)-\sin \phi=\sin (2\phi) $ I tried to expand $\sin (5\phi)$ and $\sin (2\phi)$, so the equation only contains $\cos\phi$ and $\sin \phi$.

But I can't make it into a form like $\sin(\phi+a)=n$ to get the general solution.

There's another question like this :

Find the general solution of this equation $ \sin 2x+\sin 4x=\cos 2x+\cos 4x $

3 Answers 3

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I think I see a method of solving the second equation. First rewrite the equation as

$\sin2x-\cos2x=\cos4x-\sin4x$

Now square both sides

$\sin^22x-2\sin2x\cos2x+\cos^22x=\cos^24x-2\cos4x\sin4x+\sin^24x$

$1-\sin4x=1-2\cos4x\sin4x$

$\sin4x(1-2\cos4x)=0$

This form should be a lot easier to solve, though new roots may have been introduced by squaring both sides.

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$\sin 5\phi-\sin\phi = 2\cos 3\phi \sin 2\phi$, therefore, equation becomes, $2\cos 3\phi \sin2\phi = \sin 2\phi$ giving possibilities, $\sin2\phi=0$ or $\cos3\phi = 1/2$. For first equation, $2\phi=n\pi \implies \phi=n\pi/2$. For second case, $3\phi = 2n\pi + \pi/3$ or $2n\pi - \pi/3$ giving $\phi = \frac{2n\pi}{3} + \pi/9$ or $\frac{2n\pi}{3} - \pi/9$ . therefore solutions are, $\phi=n\pi/2, \frac{2n\pi}{3} + \pi/9,\frac{2n\pi}{3} - \pi/9$

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    @Vic:you can visit http://www.analyzemath.com/calculus/table/table_math_formulas.html2012-06-23
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Just for your request from avtar:

$\sin(n\theta)=\binom{n}{1}\cos^{n-1}(\theta)\sin(\theta)-\binom{n}{3}\cos^{n-3}(\theta)\sin^3(\theta)+...$

Now put $n=5$.

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    @Vic.: See above second comment.2012-06-23