The equation sin^4 x + cos^4 x + sin 2x + alp | vedclass

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(C) Given the equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x =

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$.Substituting this into the equation:$1 - \frac{1}{2}\sin^2 2x + \sin 2x + \alpha = 0$.Let $y = \sin 2x$,where $-1 \le y \le 1$.The equation becomes $1 - \frac{1}{2}y^2 + y + \alpha = 0$,which simplifies to $y^2 - 2y - 2(1 + \alpha) = 0$.For $y$ to have a real solution,the discriminant $D \ge 0$:$D = (-2)^2 - 4(1)(-2(1 + \alpha)) = 4 + 8(1 + \alpha) = 12 + 8\alpha \ge 0 \Rightarrow \alpha \ge -\frac{3}{2}$.The roots are $y = \frac{2 \pm \sqrt{4 + 8(1 + \alpha)}}{2} = 1 \pm \sqrt{1 + 2(1 + \alpha)} = 1 \pm \sqrt{3 + 2\alpha}$.Since $-1 \le y \le 1$,we must have $-1 \le 1 - \sqrt{3 + 2\alpha} \le 1$ (the root $1 + \sqrt{3 + 2\alpha}$ will always be $\ge 1$).$-1 \le 1 - \sqrt{3 + 2\alpha}$ $\Rightarrow \sqrt{3 + 2\alpha} \le 2$ $\Rightarrow 3 + 2\alpha \le 4$ $\Rightarrow \alpha \le \frac{1}{2}$.Thus,the range is $-\frac{3}{2} \le \alpha \le \frac{1}{2}$.

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

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