Dear Uncle Colin,

I just solved $2\cos^2(4x)=1$ between 0 and $2\pi$ and found four solutions: $\frac{1}{16}\pi$, $\frac{3}{16}\pi$, $\frac{5}{16}\pi$ and $\frac{7}{16}\pi$. The answer scheme says there are sixteen solutions! Where have I gone wrong?

Have You Perhaps A Trig Identity Answer?

Hi, HYPATIA, and thanks for your message!

Looking at your answer, I suspect that you’ve solved for when $4x$ is between 0 and $2\pi$, rather than when $x$ is in that interval. There’s an easy fix, though!

## Adjust the limits!

Whenever your trig argument - here, $4x$ - is different from the variable you have information about, I like to adjust the interval accordingly. If you know $0 < x < 2\pi$, then you know $0 \le 4x \le 8\pi$ and can solve accordingly.

From there, it goes just as you’ve presumably done: $\cos(4x) = \pm \frac{1}{\sqrt{2}}$; find one solution ($4x = \frac{1}{4}\pi$) and use the graph symmetries to find the other 15. I could list them, but I won’t; instead, I’ll write them in compact form, $4x = \frac{2n+1}{4}\pi$ for $n = 0,1,\dots,31$.

Lastly, you don’t want $4x$, you want $x$, so divide them all by 4: $x = \frac{2n+1}{16}\pi$ for $n=0,1,\dots,31$.

A little check that the last one is really in the interval you want - yep, it’s slightly less than $2\pi$ - and you’re good to go!

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.