Question

If $3 \sin \theta = 2 \sin 3 \theta$ and $0 < \theta < \pi$, then $\sin \theta =$

• A. $\frac{2}{5}$
• B. $\frac{3}{2\sqrt{2}}$
• C. $\frac{2}{3}$
• D. $\frac{3}{5}$

Hint:

When evaluating trigonometric equations over open intervals like $(0, \pi)$, remember that $\sin\theta$ can never be zero. This allows you to safely divide both sides of the equation by $\sin\theta$ right at the start: $3 = 2(3 - 4\sin^2\theta) \implies 3 = 6 - 8\sin^2\theta \implies 8\sin^2\theta = 3$. This bypasses full factorization and saves time!

Answer 1

The Correct Option is B

Step 1: Bring in the triple-angle identity.
Replace $\sin 3\theta$ using $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ in the equation $3\sin\theta = 2\sin 3\theta$.
Step 2: Substitute.
$3\sin\theta = 2(3\sin\theta - 4\sin^3\theta) = 6\sin\theta - 8\sin^3\theta$.
Step 3: Move everything to one side.
$8\sin^3\theta - 3\sin\theta = 0$.
Step 4: Factor out the common term.
$\sin\theta\,(8\sin^2\theta - 3) = 0$, giving $\sin\theta = 0$ or $\sin^2\theta = \dfrac{3}{8}$.
Step 5: Reject the invalid root.
Since $0 < \theta < \pi$, $\sin\theta = 0$ is not allowed (it would force $\theta = 0$ or $\pi$), so we keep $\sin^2\theta = \dfrac{3}{8}$.
Step 6: Take the positive root.
On $(0,\pi)$, $\sin\theta > 0$, so $\sin\theta = \sqrt{\dfrac{3}{8}} = \dfrac{\sqrt{3}}{2\sqrt{2}} = \dfrac{3}{2\sqrt{2}}$ in the given form, which is option 2 and agrees with the key.
\[ \boxed{\sin\theta = \dfrac{3}{2\sqrt{2}}} \]