Trigonometry
Quadratic in sinx; range constraint
MMTS_Full_Test_08
Grade 12
Question:
If a root of the equation $n^2\sin^2 x - 2\sin x - (2n+1) = 0$ lies in $\left[\dfrac{\pi}{2}, \pi\right]$, then the minimum positive integer value of $n$ is
Step-by-Step Solution
Key Concept: For $x\in[\pi/2,\pi]$: $\sin x\in(0,1]$. Solve quadratic in $\sin x$ and impose $\sin x\in(0,1]$.
Quadratic analysis gives $n\geq3$. Minimum $n=3$.
Correct Answer: 3