We recognise the left hand side to be in the form:

`sin(a − b) =` ` sin a cos b − cos a sin b,`

where `a = 2x` and `b =x`.

So

`sin 2x\ cos x − cos 2x\ sin x`

`= sin(2x − x)`

`= sin x`

Now, we know the solutions of `sin x = 0` to be:

`x = 0, π`.

[Why?]

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