Question

(sin^2x-cos^2x)^2-4cos^2xsin^2x=-0,5

Answer 1

$(sin^2x-cos^2x)^2-4cos^2xsin^2x=-1/2\\\\cos^22x-sin^22x=-1/2\\\\cos(4x)=-1/2\\\\4x=\pm arccos(-1/2)+2\pi n, n\in Z\\\\x=\pm 1/4 arccos(-1/2)+\pi/2 n, n\in Z$

$arccos(-1/2)= \pi-arccos(1/2)=\pi-\pi/3=2\pi/3$

Значит, $x=\pm \frac{\pi}{6} + \frac{\pi n}{2}, n\in Z$

Answer 2

Let's simplify the given expression step by step to see if it's equal to -0.5:

1. Start with the left-hand side (LHS) expression: (sin^2x - cos^2x)^2 - 4cos^2xsin^2x

2. Use the identity: sin^2x - cos^2x = -cos(2x) LHS = (-cos(2x))^2 - 4cos^2xsin^2x

3. Simplify further using trigonometric identities: LHS = cos^2(2x) - 4cos^2xsin^2x

4. Use the double angle identity: cos^2(2x) = (1 + cos(4x))/2 LHS = (1 + cos(4x))/2 - 4cos^2xsin^2x

5. Use the identity: sin^2x = 1 - cos^2x LHS = (1 + cos(4x))/2 - 4cos^2x(1 - cos^2x)

6. Distribute the -4cos^2x: LHS = (1 + cos(4x))/2 - 4cos^2x + 4cos^4x

7. Combine like terms: LHS = (1 + cos(4x))/2 + 4cos^4x - 4cos^2x

Now, let's set this expression equal to -0.5 and solve for x:

(1 + cos(4x))/2 + 4cos^4x - 4cos^2x = -0.5

Multiply the entire equation by 2 to eliminate the fraction:

1 + cos(4x) + 8cos^4x - 8cos^2x = -1

Rearrange to get all terms on one side:

8cos^4x - 8cos^2x + cos(4x) + 1 + 1 = 0

8cos^4x - 8cos^2x + cos(4x) + 2 = 0

Unfortunately, this equation is quite complicated, and it doesn't seem to have a simple closed-form solution for x. It might be possible to solve it numerically using approximation methods or a graphing calculator, but a precise solution might not be readily available. If you have more context or constraints on x, it may be possible to simplify or further analyze the equation.