Let S be the set of all - R such that the equation- cos2x-sin x-2-7 has a solution- Then S is equal to-

Cos2x+αsin x=2α 7 ⇒1 2sin^2x+αsin x=2α 7 ⇒ 2sin^2x αsin x+2α 8=0 ⇒sin x=α±√(α^2 8(2α 8))4 ⇒sin x=α 42,2 As sin x=2 is not possible so, sin x=α 42 ⇒ 1≤α 42≤1 ⇒ ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiple of an Angle

Let S be the ...

Question

Let $S$ be the set of all $α \in R$ such that the equation, $cos 2 x + α sin x = 2 α - 7$ has a solution. Then $S$ is equal to:

R

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[2, 6]

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[1, 4]

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[3, 7]

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Solution

The correct option is B $[2, 6]$
$cos 2 x + α sin x = 2 α - 7$
$\Rightarrow 1 - 2 {sin}^{2} x + α sin x = 2 α - 7 \Rightarrow 2 {sin}^{2} x - α sin x + 2 α - 8 = 0 \Rightarrow sin x = \frac{α \pm \sqrt{α^{2} - 8 (2 α - 8)}}{4} \Rightarrow sin x = \frac{α - 4}{2}, 2$
As $sin x = 2$ is not possible so,
$sin x = \frac{α - 4}{2} \Rightarrow - 1 \leq \frac{α - 4}{2} \leq 1 \Rightarrow 2 \leq α \leq 6$

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

The correct option is B [2,6]cos2x+αsinx=2α−7 ⇒1−2sin2x+αsinx=2α−7⇒2sin2x−αsinx+2α−8=0⇒sinx=α±√α2−8(2α−8)4⇒sinx=α−42,2 As sinx=2 is not possible so, sinx=α−42⇒−1≤α−42≤1⇒2≤α≤6

Let $S$ be the set of all $α \in R$ such that the equation, $cos 2 x + α sin x = 2 α - 7$ has a solution. Then $S$ is equal to: