Question

The equation $\sin x(\sin x+\cos x)=k$ has real solutions if and only if k is a real number such that

• A. $0\le k\le \frac{1+\sqrt{2}}{2}$
• B. $2-\sqrt{3}\le k\le 2+\sqrt{3}$
• C. $0\le k\le 2-\sqrt{3}$
• D. $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

Answer 1

The correct option is D $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

$\sin x(\sin x+\cos x)=k$

$\Rightarrow k={\sin }^{2}x+\sin x\cos x=\frac{1}{2}(2{\sin }^{2}x+\sin 2x)=\frac{1}{2}(1-\cos 2x+\sin 2x)=\frac{1}{2}[1+(\sin 2x-\cos 2x\left)\right]$

Now we know $-\sqrt{2}\le \sin 2x-\cos 2x\le \sqrt{2}$

Hence for given equation to have real solutions, $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$