The equation sin x(sin x+cos x)=k has real so | vedclass

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(D) Let $f(x) = \sin x(\sin x + \cos x) = \sin^2 x + \sin x \cos x$. Using the identities $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\sin x \cos x = \frac{

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

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(D) Let $f(x) = \sin x(\sin x + \cos x) = \sin^2 x + \sin x \cos x$. Using the identities $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\sin x \cos x = \frac{\sin 2x}{2}$, we get: $f(x) = \frac{1-\cos 2x}{2} + \frac{\sin 2x}{2} = \frac{1}{2} + \frac{1}{2}(\sin 2x - \cos 2x)$. We know that for any real $	heta$, $-\sqrt{a^2+b^2} \leq a \sin 	heta + b \cos 	heta \leq \sqrt{a^2+b^2}$. For $\sin 2x - \cos 2x$, we have $a=1, b=-1$, so $-\sqrt{1^2+(-1)^2} \leq \sin 2x - \cos 2x \leq \sqrt{1^2+(-1)^2}$, which simplifies to $-\sqrt{2} \leq \sin 2x - \cos 2x \leq \sqrt{2}$. Dividing by $2$, we get $-\frac{\sqrt{2}}{2} \leq \frac{\sin 2x - \cos 2x}{2} \leq \frac{\sqrt{2}}{2}$. Adding $\frac{1}{2}$ to all parts: $\frac{1}{2} - \frac{\sqrt{2}}{2} \leq \frac{1}{2} + \frac{\sin 2x - \cos 2x}{2} \leq \frac{1}{2} + \frac{\sqrt{2}}{2}$. Thus, $\frac{1-\sqrt{2}}{2} \leq f(x) \leq \frac{1+\sqrt{2}}{2}$. Since $f(x) = k$, the range of $k$ is $\frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$.

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

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