Question

The equation $\sin x.(\sin x+\cos x)=k$ has real solutions then:

• 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 $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

The given equation can be written as,

$\sin \left(x\right)\left(\sin \right(x)+\sqrt{1-{\sin }^{2}x})=k$
$k-{\sin }^2\left(x\right)=\sin \left(x\right)\sqrt{1-{\sin }^{2}x}$
$k^2+{\sin }^4\left(x\right)-2k{\sin }^2\left(x\right)={\sin }^2\left(x\right)-{\sin }^4\left(x\right)$
$2{\sin }^4\left(x\right)-(2k+1){\sin }^2\left(x\right)+k^2=0$
For real solution
$b^2-4ac>0$
$(2k+1{)}^2-8k^2>0$
$4k^2+4k+1-8k^2>0$
$4k^2-4k-1<0$
$(2k-1{)}^2-2<0$
$-\sqrt{2}<2k-1<\sqrt{2}$
$\therefore \frac{1-\sqrt{2}}{2}<k<\frac{1+\sqrt{2}}{2}$.