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!
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