Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

We have, ${\sin }^{4}x-(k+2){\sin }^{2}x-(k+3)=0$ $\Rightarrow {\sin }^{2}x=\frac{(k+2)\pm \sqrt{(k+2{)}^2+4(k+3)}}{2}$ $=\frac{(k+2)\pm (k+4)}{2}$ $\Rightarrow {\sin }^{2}x=k+3(\because {\sin }^{2}x=-1$ is not possible) Since $0\le {\sin }^{2}x\le 1$ $\therefore 0\le \mathrm{k}+3\le 1$ or $-3\le \mathrm{k}\le -2$