Trigonometric Equations
Number of Solutions Determines n — Roots of Quadratic
nta_pyq_2024_jan
Grade 11
Question:
If $2\sin^3x+2\sin x\cos^2x+4\sin x-4=0$ has exactly 3 solutions in the interval $\left[0,\dfrac{n\pi}{2}\right]$, $n\in\mathbb{N}$, then the roots of the equation $x^2+nx+(n-3)=0$ belong to:
$(0,\infty)$
$(-\infty,0)$
$\left(-\dfrac{\sqrt{17}}{2},\dfrac{\sqrt{17}}{2}\right)$
$\mathbb{Z}$
Step-by-Step Solution
Key Concept: Simplify: $2\sin^3x+2\sin x(1-\sin^2x)+4\sin x-4=0\Rightarrow6\sin x-4=0\Rightarrow\sin x=2/3$. Each period $[0,2\pi]$ gives 2 solutions; need exactly 3, so $n=5$. Quadratic: $x^2+5x+2=0\Rightarrow x=(-5\pm\sqrt{17})/2$. Both roots are negative.
$n=5$. $x^2+5x+2=0$: roots both negative. Roots $\in(-\infty,0)$.
Correct Answer: 2