Question

If \( \sqrt{3} \cos \theta + \sin \theta > 0 \), then the range of \( \theta \) is:

• A. $ -\frac{\pi}{2}<\theta<\frac{\pi}{2} $
• B. $ -\frac{\pi}{3}<\theta<\frac{2\pi}{3} $
• C. $ -\frac{2\pi}{3}<\theta<\frac{\pi}{3} $
• D. $ -\frac{\pi}{6}<\theta<\frac{5\pi}{6} $

Hint:

To simplify expressions involving linear combinations of sine and cosine, use identities that reduce them to a single trigonometric function. Then solve inequalities using known intervals where the function is positive or negative.

Answer 1

The Correct Option is B

Step 1: Rewrite the inequality.
Given: $$ \sqrt{3}\cos\theta + \sin\theta>0. $$ Factor out 2: $$ \sqrt{3}\cos\theta + \sin\theta = 2\left(\frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta\right). $$ Recognize: $$ \frac{\sqrt{3}}{2} = \cos\frac{\pi}{6}, \quad \frac{1}{2} = \sin\frac{\pi}{6}. $$ Use angle addition identity: $$ \cos\frac{\pi}{6}\cos\theta + \sin\frac{\pi}{6}\sin\theta = \cos\left(\theta - \frac{\pi}{6}\right). $$ So: $$ \sqrt{3}\cos\theta + \sin\theta = 2\cos\left(\theta - \frac{\pi}{6}\right). $$ Inequality becomes: $$ \cos\left(\theta - \frac{\pi}{6}\right)>0. $$ Step 2: Solve the cosine inequality.
Cosine is positive in: $$ -\frac{\pi}{2} + 2k\pi<\theta - \frac{\pi}{6}<\frac{\pi}{2} + 2k\pi. $$ Solve for $ \theta $: $$ -\frac{\pi}{2} + \frac{\pi}{6} + 2k\pi<\theta<\frac{\pi}{2} + \frac{\pi}{6} + 2k\pi. $$ Simplify: $$ -\frac{\pi}{3} + 2k\pi<\theta<\frac{2\pi}{3} + 2k\pi. $$ For principal values $ -\pi<\theta<\pi $, take $ k = 0 $: $$ -\frac{\pi}{3}<\theta<\frac{2\pi}{3}. $$ Step 3: Final Answer.
$$ \boxed{-\frac{\pi}{3}<\theta<\frac{2\pi}{3}} $$