Finding the solution for a trigonometric equation in a specified domain Solve for solutions from $0\le x<2\pi$:
$$\sec(4x)=2$$
When I solve this equation my answer comes to be $x=\pi/12,5\pi/12$.
However when I graph the equation $y=\sec(4x)-2$, there 6 other values for $x$ that equal zero.
How would I find these values for $x$ using the previous answers I found?
 A: This is a very common mistake that I see a lot among students — you applied the given constraint inappropriately. I bet your reasoning was as follows:

$\sec(4x)=2$ has only two solutions within the interval $[0,2\pi)$, which are $\pi/3$ and $5\pi/3$. So $4x=\pi/3,5\pi/3$, and dividing by $4$ we get $x=\pi/12,5\pi/12$.

Your mistake is that you effectively said that $\color{red}{4x\text{ is in } [0,2\pi)}$, but that is NOT the given requirement — the given constraint is that $\color{green}{x\text{ is in } [0,2\pi)}$. Again: in your first step you were solving for $4x$, and you chose values for $4x$ only within $[0,2\pi)$, but $4x$ is NOT required to lie within this interval. You should only apply this constraint to values of $x$, not of anything else.
Here's how I suggest solving constrained trig equations. First of all, forget about the constraint and find ALL solutions:
$$\sec(4x)=2 \quad \Longrightarrow \quad 4x=\frac{\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n \quad \Longrightarrow \quad x=\frac{\pi}{12}+\frac{\pi n}{2},\frac{5\pi}{12}+\frac{\pi n}{2}.$$
Now, that you have the values of $x$ (not of something else, like $4x$), it's time to check which ones of them lie withing the required interval. This way you will find all requested solutions to the equation. In this example, in both families $n=0,1,2,3$ produce values of $x$ within $0\le x<2\pi$, so there are $8$ solutions.
A: Well, note that we have the following:


*

*$\sec(4x)=2$ if and only if $\cos(4x)=\frac12$

*$0\leq x<2\pi$ if and only if $0\leq 4x<8\pi$


Thus, you need only find the solutions of $\cos(t)=\frac12$ for $0\leq t<8\pi,$ and convert to solutions of the given equation via $x=\frac14t.$
