Question

1. determine all possible exact values of θ where ( \tan\theta = -sqrt{3} ), and ( -2pileq\thetaleq2pi ). (2 marks)

solution(s):

Explanation:

Step1: Find the reference angle

We know that \(\tan\theta =-\sqrt{3}\). The reference angle \(\alpha\) is found by \(\tan\alpha=\sqrt{3}\), so \(\alpha = \frac{\pi}{3}\) (since \(\tan\frac{\pi}{3}=\sqrt{3}\)).

Step2: Determine the quadrants

Since \(\tan\theta=-\sqrt{3}<0\), \(\theta\) is in the second or fourth quadrant.
The general form of the solutions for \(\tan\theta =-\sqrt{3}\) is \(\theta=\frac{2\pi}{3}+n\pi\), where \(n\in\mathbb{Z}\) (because \(\tan\theta\) has a period of \(\pi\)).

Step3: Find solutions in the range \(- 2\pi\leq\theta\leq2\pi\)

When \(n = - 2\): \(\theta=\frac{2\pi}{3}-2\pi=\frac{2\pi - 6\pi}{3}=-\frac{4\pi}{3}\)
When \(n=-1\): \(\theta=\frac{2\pi}{3}-\pi=\frac{2\pi - 3\pi}{3}=-\frac{\pi}{3}\)
When \(n = 0\): \(\theta=\frac{2\pi}{3}\)
When \(n = 1\): \(\theta=\frac{2\pi}{3}+\pi=\frac{2\pi+3\pi}{3}=\frac{5\pi}{3}\)

Answer:

\(\theta=-\frac{4\pi}{3},-\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\)