Question

Solve the equation on the interval $0\le \theta \le 2\mathrm{\pi }$.

$2{\sin }^2\theta -3\sin \theta +1=0$

Short answer

On solving the equation, we get,

$\theta \in \left\{\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6}\right\}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} 2{\sin }^2\theta -3\sin \theta +1=0 \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

On rewriting the equation,

$$\begin{gathered} 2{\sin }^2\theta -2\sin \theta -\sin \theta +1=0 \\ 2\sin \theta \left(\sin \theta -1\right)-\left(\sin \theta -1\right)=0 \\ \left(\sin \theta -1\right)\left(2\sin \theta -1\right)=0 \end{gathered}$$

Then,

$$\begin{gathered} \sin \theta -1=0 \\ \sin \theta =1 \end{gathered}$$

Also,

$$\begin{gathered} 2\sin \theta -1=0 \\ \sin \theta =\frac{1}{2} \end{gathered}$$

Step 2. Solve the equation for the given interval.

Consider the given question,

$$\begin{gathered} \sin \theta =1 \\ \sin \theta =\frac{1}{2} \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

Solving $\sin \theta =1$for the given interval,

role="math" localid="1646747872475" $$\begin{gathered} \sin \theta =1 \\ \theta ={\sin }^{-1}\left(1\right) \\ \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Solving $\sin \theta =\frac{1}{2}$for the given interval,

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{1}{2}\right) \\ \theta =\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6} \end{gathered}$$

Therefore, the solution set is$\theta \in \left\{\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6}\right\}$