Question

Find all values of $\alpha$ for which the equation ${\sin }^{4}x+{\cos }^4+\sin 2x+\alpha =0$ is valid. Also find the general solution of the equation.

Answer 1

${\sin }^{4}x+{\cos }^{4}x+\sin 2x+\alpha =0$

$\therefore ({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2+\sin 2x+\alpha =0$

$\therefore ({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x.{\cos }^{2}x+\sin 2x+\alpha =0$

$1-2{\sin }^{2}x.{\cos }^{2}x+\sin 2x+\alpha =0$

$1-2\frac{{\sin }^{2}2x}{4}+\sin 2x+\alpha =0$

$2-{\sin }^{2}2x+2\sin 2x+2\alpha =0$

$2\alpha ={\sin }^{2}2x-2{\sin }^{2}x-2$

let $\sin 2x=y$

$2\alpha =y^2-2y-2$

$2\alpha =(y-1{)}^2-1-2$

$2\alpha =(y-1{)}^2-3$

$\alpha =\frac{(y-1{)}^2}{2}-\frac{3}{2}$

$\Rightarrow -1\le y\le 1$

$\Rightarrow -2\le y-1\le 0$

$\Rightarrow 0\le (y-1{)}^2\le 4$

$\Rightarrow -3\le (y-1{)}^2-3\le 1$

$\Rightarrow -\frac{3}{2}\le \alpha \le \frac{1}{2}$