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With addition formulae and double angle formulae,solve this equation

$ \sin(3x) - 2\sin(4x) + \sin(5x) = 0 $

For all angles between 0 degrees and 180 degrees inclusive.

Okay,Firstly I know that I can break

$\sin(3x) $ into $ \sin(2x+x)$ and just use the double angle formulae to make things easy.

But when the question adds in $\sin(5x) $ it becomes tedious even when breaking up $5x$ into $3x+2x$. So I was wondering, to reduce carelessness and to make things easier,if there was any other ways to tackle the question using addition formulae, double angle formulae, binomial theorem or any easy to understand ways.

4 Answers 4

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By the complex definition of the sine, with $z:=e^{ix}$, the equation is

$$\frac{z^3-z^{-3}-2z^4+2z^{-4}+z^5-z^{-5}}{2i}=0$$ or $$z^8-z^2-2z^9+2z+z^{10}-1=(z^8-1)(z-1)^2=0.$$

The solutions are thus $z=e^{i0}$ and the eighth roots of the unit, $z=e^{ik\pi/4}$. ($x=0$ is a triple root and there are four others in the requested range.)

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    Sorry...even though I know what's complex numbers and so on, my teachers haven't taught it yet so I can't use it at homework or even at 'o' levels... – 2017-02-02
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    @ArcNeoepi: see my other answer. – 2017-02-02
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Use $$ \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} $$ with $\alpha=5x$ and $\beta=3x$.

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    Can this formulae be derived from addition formulae?It looks much easier to solve if I can use this formulae. – 2017-02-02
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    @ArcNeoepi Yes, they can: consider $\sin(A+B)+\sin(A-B)=2\sin A\cos B$; now set $A+B=\alpha$ and $A-B=\beta$. – 2017-02-02
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    Thank you! I see that's very interesting! – 2017-02-02
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$$\sin(4x-x)-2\sin(4x)+\sin(4x+x)=$$

$$\sin(4x)\cos(x)-\cos(4x)\sin(x)-2\sin(4x)+\sin(4x)\cos(x)+\cos(4x)\sin(x)=$$

$$\sin(4x)(2\cos(x)-2).$$

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    Thanks. This made things much easier. – 2017-02-02
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HINT:

As $4x-3x=5x-4x$

write $3x=4x-x,5x=4x+x$

and use $\sin(A\pm B)$ formula