Answer to Question #235813 in Trigonometry for Ahmed

Question

Find the smallest value of parameter alpha such that equation

(sin(x))^2 * cos(2x) + alpha*(cos(x)^4-sin(x)^4)= -10(2* alpha +1)^2

has at least one real solution.

Answer 1

$$\cos^4x-\sin^4x=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)$$

$$=\cos(2x)\cdot1=\cos(2x)$$

$$\sin^2x=\dfrac{1-\cos(2x)}{2}$$

$$\dfrac{1-\cos(2x)}{2}(\cos(2x))+\alpha\cos(2x)=-10(2\alpha+1)^2$$

$$\cos^2(2x)-(1+2\alpha)\cos(2x)-20(2\alpha+1)^2=0$$

$$-1\leq \cos(2x)\leq 1$$

$$D=(-(1+2\alpha))^2-4(1)(-20(2\alpha+1)^2)=81(2\alpha+1)^2$$

$$\cos(2x)=\dfrac{(1+2\alpha)\pm9(1+2\alpha)}{2}$$

$$\cos(2x)=-4(1+2\alpha)\text{ or } \cos(2x)=5(1+2\alpha)$$

$$-1\leq-4(1+2\alpha)\leq1$$

$$-\dfrac{1}{4}\leq1+2\alpha\leq\dfrac{1}{4}$$

$$-\dfrac{5}{4}\leq2\alpha\leq -\dfrac{3}{4}$$

$$-\dfrac{5}{8}\leq\alpha\leq -\dfrac{3}{8}$$

$$-1\leq5(1+2\alpha)\leq1$$

$$-\dfrac{1}{5}\leq1+2\alpha\leq\dfrac{1}{5}$$

$$-\dfrac{6}{5}\leq2\alpha\leq -\dfrac{4}{5}$$

$$-\dfrac{3}{5}\leq\alpha\leq -\dfrac{2}{5}$$

$$-\dfrac{5}{8}=-\dfrac{25}{40}, -\dfrac{3}{5}=-\dfrac{24}{40}$$

The smallest value of parameter $\alpha$ is $-\dfrac{5}{8}$ .