Let S be the set of all α ∈ R such that the equation, cos2x + α sinx = 2α

Engineering

Mathematics

General Solution of Trigonometric Equation

Question

Let S be the set of all α ∈ R such that the equation, cos²x + α sinx = 2α – 7 has a solution. Then S is equal to :

[3, 7]

[2, 6]

[1, 4]

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Given cos²x + α sin x = 2α – 7

⇒ 1 – 2sin²x + α sin x = 2α – 7

⇒ 2sin²x – α sin x + 2α – 8 = 0

$\Rightarrow \sin x = \frac{a \pm \sqrt{α^{2} - 8 (2 α - 8)}}{4}$

$\Rightarrow \sin x = \frac{α \pm (α - 8)}{4}$

$\Rightarrow \sin x = \frac{α + α - 8}{4} . \frac{α - α + 8}{4}$

Sin x =2 (Not possible)

For solution

$- 1 \leq \frac{2 α - 8}{4} \leq 1$

$\Rightarrow - 4 \leq 2 α \leq 12$

⇒ a $\in$ [2, 6]