Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
The complete set of values of $x$ satisfying $\frac{2\sin 6x}{\sin x - 1} < 0$ and $\sec^2 x - 2\sqrt{2}\tan x \leq 0$ in $\left[0, \frac{\pi}{2}\right)$ is $[a, b) \cup (c, d]$, then find the value of $\left(\frac{cd}{ab}\right)$
Step-by-Step Solution
Key Concept: Rational inequalities require separate analysis of numerator sign, denominator sign, and domain restrictions.
The inequality $\frac{2\sin 6x}{\sin x - 1} 0$ and $\sin x < 1$. Since $\sin x = 1$ at $x = \frac{\pi}{2}$, we need $\sin x \neq 1$. The solution is $x \in (0, \frac{\pi}{6}) \cup (\frac{\pi}{3}, \pi)$.
Correct Answer: 6