Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The general value of \(\theta\) satisfying \(\sin^2\theta + \sin\theta = 2\) is</p>
<p>(a) \(n\pi + (-1)^n \frac{\pi}{6}\)</p>
<p>(b) \(2n\pi + \frac{\pi}{4}\)</p>
<p>(c) \(n\pi + (-1)^n \frac{\pi}{2}\)</p>
<p>(d) \(n\pi + (-1)^n \frac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a quadratic in $\sin\theta$ and determine which roots are valid by checking the range of sine.
<p>Setting $\sin\theta = y$, we get $y^2 + y - 2 = 0$, which factors as $(y+2)(y-1)=0$. Since $-1 \leq \sin\theta \leq 1$, only $\sin\theta = 1$ is valid, giving $\theta = n\pi + (-1)^n \frac{\pi}{2}$.</p>
Correct Answer: c