Dear Uncle Colin,

How do I solve $\sin(3x) = \sin(5x)$ for $0 \le x \lt 360$?

Seems I Need Extra Smarts

Hi, SINES, and thanks for your message!

This involves a formula I always have to look up, or work out from scratch. Today I am in a working out from scratch sort of mood.

The key thing is to rewrite what we’re trying to solve based around the *mean* of $3x$ and $5x$ - which is, of course, $4x$. Moving it all to one side:

$\sin(4x - x) - \sin(4x + x) = 0$

We can expand those:

$\sin(4x)\cos(x) - \cos(4x)\sin(x) - \left( \sin(4x)\cos(x) + \cos(4x)\sin(x)\right) = 0$

The $\sin(4x)\cos(x)$ terms cancel out and we’re left with:

$-2\cos(4x)\sin(x) = 0$

Now we’re cooking!

### Two factors, separately

Either $\sin(x) = 0$, which gives simple answers of 0 and 180 degrees, or $\cos(4x) =0$, which is somewhat trickier.

Since $0 \le x \lt 360$, we have $0 \le 4x \lt 1440$, and in that interval, $\cos(4x) =0$ at $4x = 90, 270, 450, 630, 810, 990, 1170$ and $1330$ degrees.

Mapping back to $x$, we get $x = 22.5, 67.5, 112.5, 157.5, 202.5, 247.5, 292.5,$ and $337.5$ degrees.

(A quick sanity check: $\cos(x) = 0$ has two solutions over the circle, so $\cos(4x) = 0$ should have eight. We’re good.)

That gives us a total of ten solutions.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.