Question

Find the value of $\alpha$ for which the equation $si{n}^{4}x+co{s}^{4}x+sin2x+\alpha =0$ is valid,. Also find the general solution of the equation.

Answer 1

${({\sin }^{2}x+{\cos }^{2}x)}^2-2\sin 2x.{\cos }^{2}x+\sin 2x+a=0$
$1-\frac{1}{2}{\sin }^{2}2x+\sin 2x+a=0$ or ${\sin }^{2}2x-2\sin 2x-2(1+a)=0$
$\therefore \sin 2x=\frac{2\pm \sqrt{4+8(1+a)}}{2}=1\pm \sqrt{2a+3}$ ...(1)
But $\sin 2x$ is real so $2a+3\ge 0$ i.e $a\ge \frac{-3}{2}$
Also $-1\le \sin 2x\le 1$
$-1\le 1\pm \sqrt{2a+3}\le 1$
As $-1\le \sin 2x\le 1$
So $-1\le 1\pm \sqrt{2a+3}\le 1$
Also $-1\le 1-\sqrt{2a+3}\le 1\Rightarrow 0\le \sqrt{2a+3}\le 2$
$\Rightarrow a\in [\frac{-3}{2},\frac{1}{2}]$
From (1) $\sin 2x=1-\sqrt{2a+3}$ where $a\in [\frac{-3}{2},\frac{1}{2}]$
$\therefore 2x=n\pi +{(-1)}^n\theta$ and $\sin \theta =1-\sqrt{2a+3}$
$\therefore x=\frac{n\pi }{2}+{(-1)}^n{\sin }^{-1}(1-\sqrt{2a+3})$ where $a\in [-\frac{3}{2},\frac{1}{2}]$