Ask Uncle Colin: Solving Trigonometric Equations
Dear Uncle Colin,
I’m trying to solve $2\cos(3x) = \sqrt{2}$, for $0 \leq x \lt 2\pi$, but the answers I find are outside the specified interval, and obviously I miss the ones that are in the interval. How would you tackle this?
 Like A Puzzle, Like A Cosine Equation
Hi, LAPLACE, and thanks for your message!
Quite a lot of Ask Uncle Colin posts have a sense of “There’s More Than One Way To Do It”. This one is not one of those: I am here to lay down The One True Way, the way that finally made it work for me, and the one I use with any student who doesn’t have their own reliable way.
Step 1: Make the trig function equal to a number
In some cases, this involves factorising, but here it doesn’t: we just need to say $\cos(3x) = \frac{\sqrt{2}}{2}$.
Step 2: Adjust your interval if necessary
The cosine argument here is $3x$, but the interval is given for $x$. In this case, we’re simply going to multiply everything by 3 to say $0 \leq 3x \lt 6\pi$ (that is, we’ll need three complete periods of the cosine graph.) I would also let $\theta = 3x$ to save on a bit of confusion.
Step 3: Sketch the graph
Here, you’re sketching the graph of $y = \cos(\theta)$, with $0 \leq \theta \lt 6\pi$.
Your sketch doesn’t need to be perfect, it just needs to wibble up and down in roughly the right places  as long as it shows the turning points and crossing points clearly enough for you to see the symmetries, it’s a good enough sketch.
A quick aside on how I sketch this graph

Draw $x$ and $y$axes, labelling the $x$axis every $\pi$. (For cosine, these will be the turning points; for a sine graph, they would be the crossing points).

Mark on those turning points and the crossing points midway between them.

Join up the dots with a nice smooth curve.

Hastily label the axes before @realityminus3 notices.
You can also draw on the value you’re looking for: $y = \frac{\sqrt{2}}{2}$, and mark where it crosses the curve.
Step 4: Use your brain or (if you must) your calculator to find a value
$\arccos\br{  \frac{\sqrt{2}}{2}} = \frac{3}{4}\pi$, which (in this case) corresponds to the leftmost crossing.
Step 5: Use the symmetries of the graph to find the other values.
The first value is $\piby 4$ below $\pi$, and the next must be $\piby 4$ above it, which gives $\frac{5}{4}\pi$.
The graph repeats every $2\pi$, so the remaining solutions for $\theta$ are $\frac{11}{4}\pi$, $\frac{13}{4}\pi$, $\frac{19}{4}\pi$ and $\frac{21}{4}\pi$.
Step 6: Move the solutions back into the original domain.
We made $\theta$ up  it’s equal to $3x$. So we have $3x = $ (all of those multiples of $\pi$.)
To find $x$, we just need to divide them all by 3, which gives us $x = \frac{3}{12}\pi$, $\frac{5}{12}\pi$, $\frac{11}{12}\pi$, $\frac{13}{12}\pi$, $\frac{19}{12}\pi$ and $\frac{21}{12}\pi$. (The first and last of those can be simplified to $\piby 4$ and $\frac{7}{4}\pi$, respectively.)
All six solutions are back in the interval we wanted.
(After doing enough of these, you can start to generate and remember shortcuts. In my experience, having a reliable method you trust and understand beats any number of badlymemorised rules.)
Hope that helps!
 Uncle Colin