Question

In Problems 57– 80, solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$

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

Short answer

The solution set of the given equation is$\left\{\frac{3\mathrm{\pi }}{2}\right\}$

Step-by-step solution

Step 1. Given Information 

In the given problems we have to solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$

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

Step 2. The equation in its present form contains a cosine and sine.  

However, a form of the Pythagorean Identity, ${\sin }^2\theta +{\cos }^2\theta =1$, can be used to transform the equation into an equivalent one containing only sines.

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

Add 1 and subtract ${\sin }^2\theta$on both side

$$\begin{gathered} 4+4\sin \theta -1+{\sin }^2\theta =1-{\sin }^2\theta -1+{\sin }^2\theta \\ 3+4\sin \theta +{\sin }^2\theta =0 \\ {\sin }^2\theta +4\sin \theta +3=0 \end{gathered}$$

Step 3. Now the equation is sin2θ+4sinθ+3=0

Factor the equation.

$$\begin{gathered} {\sin }^2\theta +(3+1)\sin \theta +3=0 \\ {\sin }^2\theta +3\sin \theta +\sin \theta +3=0 \\ \sin \theta (\sin \theta +3)+1(\sin \theta +3)=0 \\ (\sin \theta +1)(\sin \theta +3)=0 \end{gathered}$$

Step 4. Use the Zero-Product Property. 

$$\begin{gathered} \sin \theta +1=0\mathrm{or}\sin \theta +3=0 \\ \sin \theta +1-1=0-1\mathrm{or}\sin \theta +3-3=0-3 \\ \sin \theta =-1\mathrm{or}\sin \theta =-3 \end{gathered}$$

Step 5. Solving each equation in the interval [0,2π], we obtain  

$\theta =\frac{3\mathrm{\pi }}{2}\mathrm{or}\mathrm{No}\mathrm{solution}$

The solution set is$\left\{\frac{3\mathrm{\pi }}{2}\right\}$.