Find $\theta$ with $\sec(3\theta/2)=-2$ on the interval $[0, 2\pi]$. I started off with $\cos(3 \theta/2)=-1/2$, thus $3\theta/2 = 2\pi/3$, but I don't know what to do afterwards, the answer should be a huge list of $\theta$s, which I cannot seem to get.
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$\begingroup$A helpful technique for solving $\cos(3\theta/2) = -1/2$ is to think of $3\theta/2$ as some new variable, call it $X = 3\theta/2$. Now we're solving $\cos(X) = -1/2$, but we have to pay attention to the original bounds,
$$0 \leq \theta \le 2\pi.$$
We need to relate them to our new variable, $X = 3\theta/2$. We can do that by multiplying all three "sides" by $3/2$:
\begin{align*} 0 &\le \theta \le 2\pi \\ 0 &\le \frac{3}{2}\theta \leq \frac{3}{2}2\pi\\ 0 &\le X \le 3\pi. \end{align*}
So now you're just solving $\cos(X) = -1/2$ for $0 \leq X \leq 3\pi$. Thinking graphically,
we know that, like you said, $X = 2\pi/3$. We can see though that also $X = 4\pi/3$ will work, as well as going around one full revolution beyond $2\pi/3$ (note that $4\pi/3$ plus a full revolution of $2\pi$ won't work, since $4\pi/3$ is more than $\pi$, and another revolution would take you past $3\pi$).
Don't forget to convert your $X$s back into $\theta$s, where $\theta = \frac{2}{3}X$, since $X = \frac{3}{2}\theta$.
$\endgroup$ 3 $\begingroup$$$\sec\left(\frac{3\theta}{2}\right)=-2 \implies\cos\left(\frac{3\theta}{2}\right)=-\frac{1}{2} \implies \cos\left(\frac{3\theta}{2}\right)=\cos\left(\frac{2\pi}{3}\right)$$ Thus, we have a general solution as follows$$\frac{3\theta}{2}=2n\pi \pm \frac{2\pi}{3} \implies \theta=\frac{2}{3}\left(2n\pi\pm\frac{2\pi}{3}\right)=\frac{4\pi}{9}\left(3n\pm1 \right)$$Where, $n$ is any integer. All the values of $\theta$ in the given interval $[0, 2\pi]$ can be easily determined by setting different positive values of the integer $n=0, 1, 2, ..$ & selecting the correct values under given constraints, we get following three values of $\theta$ $$\theta=\frac{4\pi}{9}, \frac{8\pi}{9}, \frac{16\pi}{9}$$
$\endgroup$ $\begingroup$Write the equation as $$\cos\left(\frac {3\theta}{2}\right)=-\frac 12$$
The principal value (PV) is $\cos^{-1}\left(-\frac 12\right)=\frac{2\pi}{3}$
Then use the general solution for cosine to generate the values for the interval you are solving for.
The general solution is $\pm PV+n*2\pi$, where $n$ is any integer.
So we have $$\frac{3\theta}{2}=\pm \frac{2\pi}{3}+n*2\pi$$
Then $$\theta=\pm \frac{4\pi}{9}+n*\frac{4\pi}{3}$$
Now choose your integer values of $n$ so that the values are in your range and exhaustive.
$\endgroup$ $\begingroup$Your procedure is correct, except you forgot to include the many other values of $\cos^{-1}(-\frac12)$.
$\cos{x} = -\frac12$ for $x = \frac{2\pi}{3}\pm2\pi n, \frac{4\pi}{3}\pm2\pi n$, where $n$ is an integer.
So you would have:
$$ \frac{3\theta}{2} = \frac{2\pi}{3} \to \theta = \frac{4\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{8\pi}{3} \to \theta = \frac{16\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{14\pi}{3} \to \theta = \frac{28\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{20\pi}{3} \to \theta = \frac{40\pi}{9} $$
etc.
AND:
$$ \frac{3\theta}{2} = \frac{4\pi}{3} \to \theta = \frac{8\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{10\pi}{3} \to \theta = \frac{20\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{16\pi}{3} \to \theta = \frac{32\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{22\pi}{3} \to \theta = \frac{44\pi}{9} $$
etc.
The full list of solutions would be:
$$\theta = \frac{4\pi}{9}\pm\frac{4\pi}{3}$$
$$\theta = \frac{8\pi}{9}\pm\frac{4\pi}{3}$$
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