Question

The equation $\sin x(\sin x+\cos x)=k$ has real solutions if and only if $k$ is 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 $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

$\sin x(\sin x+\cos x)=k$
$\Rightarrow 2{\sin }^{2}x+2\sin x\cos x=2k$
$\Rightarrow 1-\cos 2x+\sin 2x=2k$
$\Rightarrow \sin 2x-\cos 2x=2k-1$

The equation is true if
$-\sqrt{(1{)}^2+(-1{)}^2}\le 2k-1\le \sqrt{(1{)}^2+(-1{)}^2}$
$\Rightarrow \frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$