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Question

Solve each equation on the interval $0 \leq \theta<2 \pi$. $\tan \theta+\sqrt{3}=\sec \theta$

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 8First, we simplify the equation by expressing it in terms of sines and cosines.

$\begin{aligned} \tan \theta + \sqrt{3} &= \sec \theta \\ \tan \theta + \sqrt{3} - \sec \theta &= 0\\ \frac{\sin \theta}{\cos\theta} + \sqrt{3} - \frac{1}{\cos \theta} &= 0\\ \frac{\sin \theta-1}{\cos\theta} + \sqrt{3} &= 0\\ \frac{\sin \theta-1}{\cos\theta} + \sqrt{3} &= 0\\ \frac{\sin \theta-1}{\cos\theta} + \frac{\sqrt{3}\cos\theta}{\cos\theta} &= 0\\ \frac{\sin \theta-1+\sqrt{3}\cos\theta}{\cos\theta} &= 0\\ \end{aligned}$

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