The equation $sin x (sin x + cos x) = k$ has real solutions, where $k$ is a real number. Then

0 \leq k \leq \frac{1 + \sqrt{2}}{2}

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2 - \sqrt{3} \leq k \leq 2 + \sqrt{3}

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0 \leq k \leq 2 - \sqrt{3}

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\frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}

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Solution

The correct option is C $\frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}$
Given : $sin x (sin x + cos x) = k$
$⟹ {sin}^{2} x + sin x cos x = k$
$⟹ \frac{1 - cos 2 x}{2} + \frac{sin 2 x}{2} = k$
$⟹ 1 - cos 2 x + sin 2 x = 2 k$
$⟹ sin 2 x - cos 2 x = 2 k - 1$

$⟹ - \sqrt{2} \leq 2 k - 1 \leq \sqrt{2}$ ...... $[∵ - \sqrt{2} \leq sin 2 x - cos 2 x \leq \sqrt{2}]$

$⟹ 1 - \sqrt{2} \leq 2 k \leq 1 + \sqrt{2}$

$⟹ \frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}$

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