First slide

Trigonometric Identities

Question

If $\frac{y + 3}{2 y + 5} = \sin^{2} x + 2 \cos x + 1$ then the value of y lies in the interval

Moderate

A

$(- \infty, - \frac{8}{3}]$
B

$[- \frac{12}{5}, \infty)$
C

$[- \frac{8}{3}, - \frac{12}{5}]$
D

$[- \frac{8}{3}, \infty)$

Solution

We have
$\begin{array}{l} \frac{y + 3}{2 y + 5} = \sin^{2} x + 2 \cos x + 1 \\ ∴ \cos^{2} x - 2 \cos x + 1 = 3 - \frac{y + 3}{2 y + 5} \\ ∴ (\cos x - 1)^{2} = \frac{5 y + 12}{2 y + 5} \end{array}$
Now $- 1 \leq \cos x \leq 1$
$\begin{array}{ll} ∴ - 2 \leq \cos x - 1 \leq 0 \\ ∴ 0 \leq (\cos x - 1)^{2} \leq 4 \\ ∴ 0 \leq \frac{5 y + 12}{2 y + 5} \leq 4 \\ ∴ 0 \leq \frac{5 y + 12}{2 y + 5} and \frac{5 y + 12}{2 y + 5} \leq 4 \\ ∴ \frac{5 y + 12}{2 y + 5} \geq 0 and \frac{3 y + 8}{2 y + 5} \geq 0 \\ ∴ (- \infty, - \frac{5}{2}) \cup [- \frac{12}{5}, \infty) and (- \infty, - \frac{8}{3}) \cup [- \frac{5}{2}, \infty) \\ ∴ y \in (- \infty, - \frac{8}{3}] \cup [- \frac{12}{5}, \infty) \end{array}$

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