The equation {sin ^4}x + {cos ^4}x + sin 2x + | vedclass

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Question

(c) ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$

==> ${({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$

==&gt

Vedclass · Accepted Answer

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

Vedclass · Answer

(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}$.

The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for

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