Trigonometrical Equations 1 Question 1
1. Let $S$ be the set of all $\alpha \in R$ such that the equation, $\cos 2 x+\alpha \sin x=2 \alpha-7$ has a solution. Then, $S$ is equal to
(2019 Main, 12 April II)
(a) $R$
(b) $[1,4]$
(c) $[3,7]$
(d) $[2,6]$
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Answer:
Correct Answer: 1. (d)
Solution:
- The given trigonometric equation is
$$ \begin{array}{lc} & \cos 2 x+\alpha \sin x=2 \alpha-7 \\ \Rightarrow & 1-2 \sin ^{2} x+\alpha \sin x=2 \alpha-7 \\ & \quad\left[\because \cos 2 x=1-2 \sin ^{2} x\right] \\ \Rightarrow & 2 \sin ^{2} x-\alpha \sin x+2 \alpha-8=0 \\ \Rightarrow & 2\left(\sin ^{2} x-4\right)-\alpha(\sin x-2)=0 \\ \Rightarrow & 2(\sin x-2)(\sin x+2)-\alpha(\sin x-2)=0 \end{array} $$
$$ \begin{aligned} & \Rightarrow \quad(\sin x-2)(2 \sin x+4-\alpha)=0 \\ & \therefore \quad 2 \sin x+4-\alpha=0 \\ & \Rightarrow \quad \sin x=\frac{\alpha-4}{2} \end{aligned} $$
$[\because \sin x-2 \neq 0]$
Now, as we know $-1 \leq \sin x \leq 1$
$$ \begin{array}{ll} \therefore & -1 \leq \frac{\alpha-4}{2} \leq 1 \\ \Rightarrow & -2 \leq \alpha-4 \leq 2 \Rightarrow 2 \leq \alpha \leq 6 \Rightarrow \alpha \in[2,6] \end{array} $$
