Trigonometric Equations
Transcendental Equation — Counting Solutions
nta_pyq_2024_apr
Grade 11
Question:
The number of solutions of $\sin^2x+(2+2x-x^2)\sin x-3(x-1)^2=0$, where $-\pi\leq x\leq\pi$, is _____
Step-by-Step Solution
Key Concept: Rewrite: $\sin^2x-(x-1)^2\sin x-3(x-1)^2=0$. Treat as quadratic in $\sin x$: $(\sin x+3)(\sin x-(x-1)^2)=... $ factoring: $[\sin x+3][\sin x-(x-1)^2]=0$... actually factor as $(\sin x-(x-1)^2)(\sin x+3)=0$.
$\sin x=(x-1)^2$ has 2 solutions in $[-\pi,\pi]$.
Correct Answer: 2