[Math] How to solve $\tan^2 \theta − 2 \sec \theta = 2$

trigonometry

Solve the given equation. Find all solutions of the equation (express your answer in terms of k, where k is any integer).

$$\tan^2 \theta − 2 \sec \theta = 2$$

What do I do to solve for $\theta$?

update: I got

$cos\theta$ $= \frac {1}{3}$ and $cos\theta$ $-1$

and then:

$1.231+2\pi k,\space5.052+2\pi k,\space\pi +2\pi k$

Thank you!

Best Answer

Notice, we have $$\tan^2\theta-2\sec\theta=2$$ $$\sec^2\theta-1-2\sec\theta=2$$ $$\sec^2\theta-2\sec\theta-3=0$$ $$\sec^2\theta-3\sec\theta+\sec\theta-3=0$$ $$\underbrace{\sec^2\theta-3\sec\theta}\underbrace{+\sec\theta-3}=0$$ $$\sec\theta(\sec \theta-3)+(\sec\theta-3)=0$$ $$(\sec\theta-3)(\sec\theta+1)=0$$ $$\implies \sec\theta-3=0\iff \cos \theta=\frac{1}{3}\iff \color{blue}{\theta=2k\pi\pm\cos^{-1}\left(\frac{1}{3}\right)}$$

$$\implies \sec\theta+1=0\iff \cos \theta=-1\iff \color{blue}{\theta=2k\pi+\pi}$$

Where, $k$ is any integer & $\cos^{-1}\left(\frac{1}{3}\right)=1.231$

Your answers are obtained as follows, $$\color{red}{\theta}=2k\pi+\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+1.231}$$ $$\color{red}{\theta}=2k\pi+2\pi-\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+5.052}\ \ \ \text{(since, }\ \cos\theta=\cos(2\pi-\theta))$$ & $$\color{red}{\theta=2k\pi+\pi}$$