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find the $x$ :

$$\sin x+\sin 2x+\sin 4x=0$$

My Try :

$$\sin x+\sin 2x+\sin 4x=0 \\ ×2\sin \left(\frac{d}{2}\right)\neq0 \\ \left(2\sin \left(\frac{d}{2}\right)\right)\sin x+\left(2\sin \left(\frac{d}{2}\right)\right)\sin 2x +\left(2\sin \left(\frac{d}{2}\right)\right)\sin 4x=0 \\ 2\sin p\sin q=\cos(p-q)-\cos(p+q) \\ \cos\left(\frac{d}{2}-x\right)-\cos\left(\frac{d}{2}+x\right)+\cos\left(\frac{d}{2}-2x\right)-\cos\left(\frac{d}{2}+2x\right)+\cos\left(\frac{d}{2}-4x\right)-\cos\left(\frac{d}{2}+4x\right)=0$$

Now ?

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4 Answers

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expanding all Terms we obtain: $$\sin \left( x \right) -2\,\sin \left( x \right) \cos \left( x \right) +8\,\sin \left( x \right) \left( \cos \left( x \right) \right) ^{3} =0$$ now can we factorize as follows: $$\sin \left( x \right) \left( 8\, \left( \cos \left( x \right) \right) ^{3}-2\,\cos \left( x \right) +1 \right) =0$$ can you finish?

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HINT...Use double angle formulas to write it as $$\sin x+2\sin x\cos x+2(2\sin x\cos x)\cos2x=0$$

From this we have either $\sin x=0$ or a cubic equation in $\cos x$ obtained from using $\cos 2x=2\cos^2x-1$ which has one real solution...

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We use the double angle formulae to rewrite the equation:

\begin{align} \sin x+\sin (2x)+\sin (4x)&=0\\ \sin x+2\sin x\cos x+2\sin(2x)\cos (2x)&=0\\ \sin x+2\sin x\cos x+(4\sin x\cos x)\cos (2x)&=0\\ \sin x +2\sin x \cos x + 4\sin x \cos x\left(2\cos^2x-1\right)&=0\\ \sin x\left(1+2\cos x+4\cos x\left(2\cos^2x-1\right)\right)&=0\\ \sin x\left(1+2\cos x+4\cos x\left(2\cos^2x-1\right)\right)&=0\\ \sin x\left(1+2\cos x+8\cos^3x-4\cos x\right)&=0\\ \sin x\left(8\cos^3x-2\cos x+1\right)&=0\end{align}

And therefore we have $2$ options, $\sin x=0$ or $\left(8\cos^3x-2\cos x+1\right)=0$

The first is trivial: $\sin x=0\implies x=n\pi$, $n\in \mathbb{Z}$

The second is a cubic equation in $\cos x$:

\begin{align}8\cos^3x-2\cos x+1&=0\\ 8t^3-2t+1&=0\end{align}

We can solve this to find up to three values of $t$ (hint: we can plot the graph and see that there will only be one value of $t$ for this equation), and then solve $t=\cos x$ for each of them found.

Can you continue from here?

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sinx + sin2x + sin4x = 0

sinx + 2sin3x . sinx = 0

sinx (1 +2sin3x) = 0

sinx = 0 or sin3x = -1/2

x = npi or x = npi + -1^n (7pi/18)

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