How do you solve \[\sin 3\theta = 1\]?

Hint: In order to determine the solution of the above trigonometric function, take the inverse of sine on both sides, derive an angle whose sine is equal to one. Generalise the solution by considering the fact that the function sine repeats itself after every $ 2\pi $ interval. Simplify the solution for $ \theta $ to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation \[\sin 3\theta = 1\].
\[\sin 3\theta = 1\]
Taking inverse of sine one both sides of the equation , we get
\[{\sin ^{ - 1}}\left( {\sin 3\theta } \right) = {\sin ^{ - 1}}\left( 1 \right)\]
Since $ {\sin ^{ - 1}}\left( {\sin } \right) = 1 $ as they both are inverse of each other
 $ 3\theta = {\sin ^{ - 1}}\left( 1 \right) $
The value of $ 3\theta $ is equal to an angle whose sine value is equal to one.
Since as we know
$ \sin \left( {\dfrac{\pi }{2}} \right) = 1 \to \dfrac{\pi }{2} = {\sin ^{ - 1}}\left( 1 \right) $ . Putting this value in our original equation, we get
 $\Rightarrow 3\theta = \dfrac{\pi }{2} $
As we know the period of sine function is $ 2\pi $ as the graph of sine function repeats itself after every $ 2\pi $ interval . So if we generalise our solution, we obtain the value of $ 3\theta $ as
 $\Rightarrow 3\theta = \dfrac{\pi }{2} + 2n\pi $ where n is any integer value
Dividing both sides of the equation with the number $ 3 $ , we have
\[
\Rightarrow \dfrac{{3\theta }}{3} = \dfrac{1}{3}\left( {\dfrac{\pi }{2} + 2n\pi } \right) \\
\Rightarrow \theta = \dfrac{\pi }{6} + \dfrac{{2\pi }}{3}n \\
 \]
So, the correct answer is “ $ \theta = \dfrac{\pi }{6} + \dfrac{{2\pi }}{3}n $ ”.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.The answer obtained should be in a generalised form.
3.The domain of sine function is in the interval $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .