Solving $ 4\sin(3\theta + 2) = 1 $

Question

Find all $ \theta \in (0, 2\pi)$ satisfying $$ 4\sin(3\theta + 2) = 1 $$

I tried using a the right triangle and I tried opening up the $\sin$ function too. But nothing seems to work.

Answer 1

We have $\sin(3\theta+2)=\frac{1}{4}$ and taking the $arcsin$ of both sides and using the CAST method (here for examples) or sketching the graph, gives the set of all solutions to be:

• $3\theta+2=2n\pi+sin^{-1}(\frac{1}{4})$
• $3\theta+2=2n\pi+\pi-sin^{-1}(\frac{1}{4})$

for $n\in\mathbb Z$.

Now you have to find all $\theta$'s in $(0,2\pi$) (you should have 6 solutions).

Answer 2

Dividing both sides by $4$, we have

$$ \sin(3\theta+2)=\frac{1}{4} $$

Then, let $y=3\theta+2$:

$$ \sin y=\frac{1}{4} $$

And use a calculator to find $\arcsin \frac{1}{4}$. This gives you the principal solution to the equation. However, we need to solve for $\theta$, not $y$, and we also need to pay attention to the range of values that $\theta$ can take:

\begin{align} 0&\leq\theta\leq2\pi \\ \implies0&\leq3\theta\leq6\pi \\ \implies2&\leq y\leq6\pi+2 \end{align}

Once you have found all the possible values of $y$, use these to find the values of $\theta$. As Wrench has pointed out, a good way to find all the possible values of $y$ is by using a CAST diagram, or by considering the symmetries of the sine graph.