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The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
$ - \frac{1}{2} \le \alpha \le \frac{1}{2}$
$ - 3 \le \alpha \le 1$
$ - \frac{3}{2} \le \alpha \le \frac{1}{2}$
$ - 1 \le \alpha \le 1$
Solution
(c) ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$
==> ${({\sin ^2}x + {\cos ^2}x)^2} – 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$
==> ${\sin ^2}2x – 2\sin 2x – 2 – 2\alpha = 0$
Let $sin 2x = y$. Then the given equation becomes
${y^2} – 2y – 2(1 + \alpha ) = 0$,
where $ – 1 \le y \le 1$, $({\rm{ }} – 1 \le \sin 2x \le 1)$
For real, discriminant
$ \ge 0$$ \Rightarrow $$3 + 2\alpha \ge 0$
$ \Rightarrow $ $\alpha \ge – \frac{3}{2}$
Also $ – 1 \le y \le 1 \Rightarrow – 1 \le 1 – \sqrt {3 + 2\alpha } \,\, \le 1$
$ \Rightarrow $ $3 + 2\alpha \le 4 \Rightarrow \alpha \le \frac{1}{2}$.
Thus $ – \frac{3}{2} \le \alpha \le \frac{1}{2}$.
