Question

solve the equation.

\\( \sin \theta + \sqrt { 3 } \cos \theta = 1 \\)

what is the solution in the interval \\( 0 \leq \theta < 2 \pi \\) ? select the correct choice and fill in any

\\( \bigcirc \\) a. \\( \theta = \\{ \\} \\)
(simplify your answer. type an exact answer, using \\( \pi \\) as needed. type your a
the expression. use a comma to separate answers as needed.)
\\( \bigcirc \\) b. there is no solution.

Explanation:

Step1: Use the formula \(a\sin\theta + b\cos\theta = c\sin(\theta+\varphi)\)

We know that \(a = 1\), \(b=\sqrt{3}\), then \(c=\sqrt{1^{2}+(\sqrt{3})^{2}} = 2\), and \(\tan\varphi=\frac{b}{a}=\sqrt{3}\), so \(\varphi=\frac{\pi}{3}\).
The equation \(\sin\theta+\sqrt{3}\cos\theta = 1\) can be rewritten as \(2\sin(\theta+\frac{\pi}{3})=1\), then \(\sin(\theta+\frac{\pi}{3})=\frac{1}{2}\).

Step2: Solve for \(\theta+\frac{\pi}{3}\)

We know that if \(\sin x=\frac{1}{2}\), then \(x = 2k\pi+\frac{\pi}{6}\) or \(x=(2k + 1)\pi-\frac{\pi}{6},k\in\mathbf{Z}\).
So \(\theta+\frac{\pi}{3}=2k\pi+\frac{\pi}{6}\) or \(\theta+\frac{\pi}{3}=(2k + 1)\pi-\frac{\pi}{6}\).

Step3: Solve for \(\theta\)

For \(\theta+\frac{\pi}{3}=2k\pi+\frac{\pi}{6}\), we have \(\theta=2k\pi+\frac{\pi}{6}-\frac{\pi}{3}=2k\pi-\frac{\pi}{6}\).
For \(\theta+\frac{\pi}{3}=(2k + 1)\pi-\frac{\pi}{6}\), we have \(\theta=(2k + 1)\pi-\frac{\pi}{6}-\frac{\pi}{3}=(2k + 1)\pi-\frac{\pi}{2}\).

Step4: Find solutions in the interval \(0\leqslant\theta<2\pi\)

When \(k = 1\) for \(\theta=2k\pi-\frac{\pi}{6}\), \(\theta=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\).
When \(k = 0\) for \(\theta=(2k + 1)\pi-\frac{\pi}{2}\), \(\theta=\pi-\frac{\pi}{2}=\frac{\pi}{2}\).

Answer:

\(\theta=\frac{\pi}{2},\frac{11\pi}{6}\)