Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
The inequality $4\sin 3x+5\geq 4\cos 2x+5\sin x$ is true for $x\in$:
$\left[-\pi,-\frac{3\pi}{2}\right]$
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
$\left[\frac{5\pi}{8},\frac{13\pi}{8}\right]$
$\left[\frac{23\pi}{14},\frac{41\pi}{14}\right]$
Step-by-Step Solution
Key Concept: Rearranging and factoring trigonometric inequalities reveals that perfect squares force constraints on the remaining factors.
From $4\sin 3x + 5 ≥ 4\cos 2x + 5\sin x$, rearranging gives $(\sin x - 1)(4\sin x + 1)^2 ≤ 0$. Since $(4\sin x + 1)^2 ≥ 0$ always, we need $\sin x - 1 ≤ 0$, which holds for all $x \in \mathbb{R}$, with equality when $\sin x = 1$.
Correct Answer: 1,2,3,4