Question

For what values of a does the Equation cos 2x + a sin x = 2a - 7 possesses a real solution?

Answer 1

Given:

• cos 2x + $\alpha$ sin x = 2$\alpha$ - 7

Solution:

$\\ \colon\implies{\tt{ cos \ 2x + \alpha \ sin \ x = 2 \alpha - 7 }} \\ \\ \\ \colon\implies{\tt{ 1 - 2 \ sin^2 x + \alpha \ sin \ x = 2\alpha- 7 }} \\ \\ \\ \colon\implies{\tt{ 2 sin^2 x - \alpha \ sin \ x + 2\alpha - 8 = 0 }}$

Discriminant,

D = (-$\alpha$ )² - 4 × 2 (2$\alpha$ - 8)

$\implies$ $\alpha$ ² - 16 $\alpha$ + 64

$\implies$ ($\alpha$- 8)²

For Real values of x,

D ≥ 0 ($\alpha$- 8)² ≥ 0

$\\ \colon\implies{\tt{ sin \ x = \dfrac{ \alpha \pm ( \alpha - 8)}{4} }} \\ \\ \\ \colon\implies{\tt{ -1 \leq sin \ x \leq 1 }} \\ \\ \\ \colon\implies{\tt{ -1 < \dfrac{ \alpha \pm ( \alpha - 8) }{4} < 1 }} \\ \\ \\ \colon\implies{\tt{ -4 \leq \alpha \pm ( \alpha - 8) \leq 4 }} \\ \\ \\ \colon\implies{\tt{ -4 \leq 2 \alpha - 8 \leq 4 }} \\ \\ \\ \colon\implies{\tt{ -4 + 8 \leq 2 \alpha \leq 4 + 8 }} \\ \\ \\ \colon\implies{\tt{ 4 \leq 2 \alpha \leq 12 }} \\ \\ \\ \colon\implies{\tt{ 2 \leq \alpha \leq 6 }}$

Hence,

• For real value of x = 2 ≤ $\alpha$ ≤ 6 .