Question

Sin 4x + sin 6x = 0 ​

Answer 1

`\sin(4x)+\sin(6x)=0`

Expand the arguments to include
`5x`:

`\sin(5x-x)+\sin(5x+x)=0`

`(\sin(5x)\cos x-\sin x\cos(5x))+(\sin(5x)\cos x+\sin x\cos(5x))=0`

`2\sin(5x)\cos x=0`

Then either

`\sin(5x)=0`

or

`\cos x=0`

In the first case, use the fact that
`\sin0=\sin\pi=0` and that
`\sin x` has period
`2\pi`, so that

`\sin(5x)=0\implies\begin{cases}5x=0+2n\pi\\5x=\pi+2n\pi\end{cases}\implies\boxed{x=\frac{2n\pi}5\text{ or }x=\frac{(2n+1)\pi}5}`

where
`n` is any integer.

In the second case, we have
`\cos\frac\pi2=\cos\frac{3\pi}2=0` and
`\cos x` also has period
`2\pi`, so that

`\cos x=0\implies\begin{cases}x=\frac\pi2+2n\pi\\\\x=\frac{3\pi}2+2n\pi\end{cases}\implies\boxed{x=\frac{(4n+1)\pi}2\text{ or }\frac{(4n+3)\pi}2}`