The equation ^ 4 x - ^ 4 x + 2x + α ^ 2 + α = 0 will have at least...

$\left(\cos^{2} x - \sin^{2} x\right)\left(\cos^{2} x + \sin^{2} x\right)$+ \cos 2x + {\alpha}^{2} + \alpha = 0$<br />$⇒ \left(\cos 2x\right)\times 1 + \cos 2x + {\alpha}^{2} + \alpha = 0$<br />$⇒ {\alpha}^{2} + \alpha = - 2\cos 2x$<br />RHS ∈ \left[- 2,2\right]$
for a solution, $- 2 \le {\alpha}^{2} + \alpha \le 2$
${\alpha}^{2} + \alpha + 2 \ge 0$ and ${\alpha}^{2} + \alpha - 2 \le 0$
Now, ${\alpha}^{2} + \alpha + 2 \ge 0$ is always true

and \left(\alpha + 2\right) \left(\alpha -1\right)$≤ 0$
$⇒ \alpha \in \left[-2,1\right]$