Additional Mathematics – Trigonometry | e-Consult

Rewrite as \(\tan\theta = \sqrt{3}\).

The basic solution of \(\tan\theta = \sqrt{3}\) is \(\theta = \frac{\pi}{3} + n\pi\), where \(n\in\mathbb Z\).

Find the values of \(n\) that keep \(\theta\) within \([-π,π]\):

• For \(n = 0\): \(\theta = \frac{\pi}{3}\) (within the interval).
• For \(n = -1\): \(\theta = \frac{\pi}{3} - \pi = -\frac{2\pi}{3}\) (within the interval).
• For \(n = 1\): \(\theta = \frac{\pi}{3} + \pi = \frac{4\pi}{3} > \pi\) (reject).

Thus the solutions are \(\displaystyle \theta = -\frac{2\pi}{3},\ \frac{\pi}{3}\).