Question

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

${\tan }^2\theta =\frac{3}{2}\sec \theta$

Short answer

The solution set of the given equation is$\left\{\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.

Step-by-step solution

Step 1. Given Information 

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

${\tan }^2\theta =\frac{3}{2}\sec \theta$

Step 2. The equation in its present form contains a secant and tangent.  

However, a form of the Pythagorean Identity, ${\sec }^2\theta -{\tan }^2\theta =1$, can be used to transform the equation into an equivalent one containing only secant.

role="math" localid="1647013734561" $({\sec }^2\theta -1)=\frac{3}{2}\sec \theta$

Multiply with 2 on both side

$$\begin{gathered} 2({\sec }^2\theta -1)=\frac{3}{2}\sec \theta \times 2 \\ 2{\sec }^2\theta -2=3\sec \theta \end{gathered}$$

Subtract $3\sec \theta$on both side

$$\begin{gathered} 2{\sec }^2\theta -2-3\sec \theta =3\sec \theta -3\sec \theta \\ 2{\sec }^2\theta -2-3\sec \theta =0 \\ 2{\sec }^2\theta -3\sec \theta -2=0 \end{gathered}$$

Step 3. Now the equation is 2sec2θ-3secθ-2=0

Factor the equation.

$$\begin{gathered} 2{\sec }^2\theta -(4-1)s\mathrm{ec}\theta -2=0 \\ 2{\sec }^2\theta -4s\mathrm{ec}\theta +s\mathrm{ec}\theta -2=0 \\ 2s\mathrm{ec}\theta (s\mathrm{ec}\theta -2)+1(s\mathrm{ec}\theta -2)=0 \\ (2s\mathrm{ec}\theta +1)(s\mathrm{ec}\theta -2)=0 \end{gathered}$$

Step 4. Use the Zero-Product Property. 

$$\begin{gathered} 2s\mathrm{ec}\theta +1=0\mathrm{or}s\mathrm{ec}\theta -2=0 \\ 2s\mathrm{ec}\theta +1-1=0-1\mathrm{or}s\mathrm{ec}\theta -2+2=0+2 \\ 2s\mathrm{ec}\theta =-1\mathrm{or}s\mathrm{ec}\theta =2 \\ \frac{2}{2}s\mathrm{ec}\theta =-\frac{1}{2}\mathrm{or}s\mathrm{ec}\theta =2 \\ s\mathrm{ec}\theta =-\frac{1}{2}\mathrm{or}s\mathrm{ec}\theta =2 \end{gathered}$$

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

$\mathrm{No}\mathrm{solution}\mathrm{or}\theta =\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}$

The solution set is$\left\{\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.