Trigonometric Equations
Trig Equations Inequations
nta_abhyas_2025
Grade 11
Question:
If $1 + \sin\theta + \sin^2\theta + \sin^3\theta + \cdots = 4 + 2\sqrt{3}, 0 < \theta < \pi$, then
\theta = \frac{\pi}{3}
\theta = \frac{\pi}{6}
\theta = \frac{\pi}{3}\text{ or }\frac{\pi}{6}
\theta = \frac{\pi}{3}\text{ or }\frac{5\pi}{6}
Step-by-Step Solution
Key Concept: Recognize infinite geometric series with ratio $\sin\theta$ and use the formula $\frac{a}{1-r}$ where $|r| < 1$ to convert the sum condition into an equation for $\sin\theta$.
Given $1 + \sin\theta + \sin^2\theta + \sin^3\theta + \cdots = 4 + 2\sqrt{3}$ with $0 < \sin\theta < 1$. Using the geometric series formula $\frac{1}{1-\sin\theta} = 4 + 2\sqrt{3}$, we get $1 - \sin\theta = \frac{1}{4 + 2\sqrt{3}} = 1 - \frac{\sqrt{3}}{2}$, so $\sin\theta = \frac{\sqrt{3}}{2}$. Thus $\theta = \frac{\pi}{3}$ or $\frac{2\pi}{3}$.
Correct Answer: 4