Questions: Solve the equation. cos θ - √3 sin θ = 1 What is the solution in the interval 0 ≤ θ < 2π?

Solution Steps

To solve the trigonometric equation \(\cos \theta - \sqrt{3} \sin \theta = 1\), we can use the method of expressing the equation in the form \(R \cos(\theta - \alpha) = 1\), where \(R\) and \(\alpha\) are constants. This involves finding \(R\) using the identity \(R = \sqrt{a^2 + b^2}\) where \(a\) and \(b\) are the coefficients of \(\cos \theta\) and \(\sin \theta\) respectively. Then, find \(\alpha\) using \(\tan \alpha = \frac{b}{a}\). Finally, solve for \(\theta\) in the given interval.

We start with the equation: \[ \cos \theta - \sqrt{3} \sin \theta = 1 \] This can be rewritten in the form \(R \cos(\theta - \alpha) = 1\).

The coefficients are \(a = 1\) and \(b = -\sqrt{3}\). We calculate \(R\) as follows: \[ R = \sqrt{a^2 + b^2} = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \] Next, we find \(\alpha\) using: \[ \tan \alpha = \frac{b}{a} = \frac{-\sqrt{3}}{1} \implies \alpha = -\frac{\pi}{3} \]

We set up the equation: \[ 2 \cos(\theta + \frac{\pi}{3}) = 1 \] Dividing both sides by 2 gives: \[ \cos(\theta + \frac{\pi}{3}) = \frac{1}{2} \] The general solutions for \(\cos x = \frac{1}{2}\) are: \[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \] Substituting back for \(\theta\):

1. \(\theta + \frac{\pi}{3} = \frac{\pi}{3} + 2k\pi \implies \theta = 2k\pi\)
2. \(\theta + \frac{\pi}{3} = -\frac{\pi}{3} + 2k\pi \implies \theta = -\frac{2\pi}{3} + 2k\pi\)

For \(k = 0\):

1. \(\theta = 0\)
2. \(\theta = -\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}\)

Thus, the solutions in the interval \(0 \leq \theta < 2\pi\) are: \[ \theta_1 = 0, \quad \theta_2 = \frac{4\pi}{3} \]

Final Answer