Question

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

$4{\sin }^2\theta =1+4\cos \theta$

Short answer

On solving the equation, we get,

$\theta \in \left\{\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} 4{\sin }^2\theta =1+4\cos \theta \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

We know, ${\sin }^2\theta =1-{\cos }^2\theta$,

$$\begin{gathered} 4\left(1-{\cos }^2\theta \right)=1+4\cos \theta \\ 4-4{\cos }^2\theta =1+4\cos \theta \\ 4{\cos }^2\theta +4\cos \theta +3=0 \\ \left(2\cos \theta -1\right)\left(2\cos \theta +3\right)=0 \end{gathered}$$

Then,

localid="1646749506085" $$\begin{gathered} 2\cos \theta -1=0 \\ \cos \theta =\frac{1}{2} \end{gathered}$$

Also,

$$\begin{gathered} 2\cos \theta +3=0 \\ \cos \theta =-\frac{3}{2} \end{gathered}$$

This equation is invalid as cosine value must lie within the range$\left[-1,1\right]$.

Step 2. Solve the equation for the given interval.

Consider the given question,

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

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

$$\begin{gathered} \cos \theta =\frac{1}{2} \\ \theta ={\cos }^{-1}\left(\frac{1}{2}\right) \\ \theta =\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3} \end{gathered}$$

Therefore, the solution set is $\theta \in \left\{\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.