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Is this statement related to trigonometry valid to write?

$$\sin\left(\frac{\pi}{6}-2x\right)=\sin4x$$

$$\implies \frac{\pi}{6}-2x=\pm 2n\pi+4x \tag{i}$$

where $n$ is any whole number.

Can I write (i)?

user
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tryingtobeastoic
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    no, because $\sin x = c$ has, in general, two solution on $[0, 2\pi]$. You will lose some solutions. The proper way is to use sum/difference to product identity to solve this. – Vasili Sep 13 '21 at 14:29

2 Answers2

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Your solution is not complete, indeed recall that

$$\sin A = \sin B \iff A=B+2k\pi \quad \lor \quad A=\pi - B+2k\pi$$

that is

$$\frac{\pi}{6}-2x=4x+2k\pi \quad\lor \quad \frac{\pi}{6}-2x=\pi-4x+2k\pi $$

Refer also to the related

user
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2

You will only find some solutions but not all of them. Using your approach, we get that $6x=\frac{\pi}{6}+2\pi n$ or $x=\frac{\pi}{36}+\frac{\pi n}{3}$.
Using sum-to-product identity, we can write $\sin (4x)+\sin(2x-\frac{\pi}{6})=0 \implies 2\sin(3x-\frac{\pi}{12})\cos(x+\frac{\pi}{12})=0$ which gives solutions $x=\frac{\pi}{36}+\frac{\pi n}{3}$ and $x=\frac{5\pi}{12}+\pi n$.

Vasili
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