Solve the given equation. (Enter your answers as a comma-separated list. Let $k$ be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
$$(\tan \theta − 2)(9 \sin^2 \theta − 1) = 0$$
So if I set both equal to zero and solve I get
$$\theta=\tan^{-1}(2)$$
$$\theta=\sin^{-1}\left(\frac13\right)$$
$$\theta=\sin^{-1}\left(-\frac13\right)$$
What do I do next?
Update:
There should be five answers, so far I have 4:
$1.107+\pi k$ which came from the tangent
$0.340+2\pi k$ which came from arcsin (1/3)
$2.802+2\pi k$ which came from arcsin (1/3)
$5.943+2\pi k$ which came from arcsin (-1/3)
There should be one more from came from arcsin (-1/3) but I don't know what to add/subtract from arcsin (-1/3) to get it.
Edit: I got the last one it was $3.481+2\pi k$
Best Answer
A Couple things:
You should get 3 results: $ (\tanθ−2)(9\sin2θ−1)=0 \Rightarrow$
$ (1) \tanθ =2$
$ (2) \sinθ = \frac{1}{3}$
$ (3) \sinθ = -\frac{1}{3} $
From here, you would use the inverse functions as you did- however, it seems as if you may use a calculator to get your results, which would probably be your best bet (assuming you are given a fixed domain)