$cos 2 x + k sin x = 2 k - 7$ has a solution for

2 \leq k \leq 6

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1 < k < 7

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4 < k < 7

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None of these

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Solution

The correct option is A $2 \leq k \leq 6$
Given, $cos 2 x + k sin x = 2 k - 7$
$\Rightarrow 1 - 2 {sin}^{2} x + k sin x = 2 k - 7$
$\Rightarrow 2 {sin}^{2} x - k sin x + 2 k - 8 - 0$
$∴ sin x = \frac{+ k \pm \sqrt{k^{2}} - 8 (2 k - 8)}{4}$
$= \frac{k \pm (k - 8)}{4}$
$\Rightarrow sin x = \frac{k - 4}{2}$
and $sin x = 2$
$∴ sin x = \frac{k - 4}{2}$
$(∵ sin x$ cannot be greater than $1$ )
Now, $- 1 \leq sin x \leq 1$
$\Rightarrow - 1 \leq \frac{k - 4}{2} \leq 1$
$\Rightarrow - 2 \leq k - 4 \leq 2$
$\Rightarrow 2 \leq k \leq 6$ .

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