Question

Sin^2 2x+ cos2x*Sin2x-2cos ^2x=0​

Answer 1

$Sin^{2} 2x+Cos2x Sin2x-2Cos^{2}2x=0|:Cos^{2}2x;Cos2x\neq0 \\\\tg^{2}2x+tg2x-2=0\\\\\left[\begin{array}{ccc}tg2x=-2\\tg2x=1\end{array}\right\\\\1)tg2x=-2\\\\2x=-arctg2+\pi n,n\in Z\\\\x=-\frac{1}{2}arctg2+\frac{\pi n }{2},n\in Z\\\\2)tg2x=1\\\\2x=arctg1+\pi n,n\in Z\\\\2x=\frac{\pi }{4}+\pi n,n\in Z\\\\x=\frac{\pi }{8}+\frac{\pi n}{2} ,n\in Z$

Answer 2

To solve the trigonometric equation sin^2(2x) + cos(2x) * sin(2x) - 2cos^2(x) = 0, we'll use trigonometric identities and algebraic manipulation. Let's break down the steps:

1. Start with the given equation: sin^2(2x) + cos(2x) * sin(2x) - 2cos^2(x) = 0

2. Use the identity sin(2x) = 2sin(x)cos(x): sin^2(2x) + 2cos(2x) * sin(x)cos(x) - 2cos^2(x) = 0

3. Use the identity cos(2x) = 2cos^2(x) - 1: sin^2(2x) + 2(2cos^2(x) - 1) * sin(x)cos(x) - 2cos^2(x) = 0

4. Distribute and simplify: sin^2(2x) + 4cos^2(x)sin(x)cos(x) - 2sin(x)cos(x) - 2cos^2(x) = 0

5. Factor out 2sin(x)cos(x) from the second and third terms: sin^2(2x) + 2sin(x)cos(x)(2cos^2(x) - 1) - 2cos^2(x) = 0

6. Use the identity 2cos^2(x) - 1 = cos(2x): sin^2(2x) + 2sin(x)cos(x)cos(2x) - 2cos^2(x) = 0

7. Rewrite sin(2x) as 2sin(x)cos(x): sin^2(2x) + 2sin(x)cos(x)cos(2x) - 2(1 - sin^2(x)) = 0

8. Distribute and simplify: sin^2(2x) + 2sin(x)cos(x)cos(2x) - 2 + 2sin^2(x) = 0

9. Combine the sin^2(x) terms: sin^2(2x) + 2sin(x)cos(x)cos(2x) + 2sin^2(x) - 2 = 0

10. Group the terms involving sin(2x): sin^2(2x) + 2sin(x)cos(x)cos(2x) + 2sin^2(x) = 2

11. Use the double angle identity for sin(2x): 2sin(x)cos(x) = sin(4x)

12. Substitute the double angle identity back into the equation: sin^2(2x) + sin(4x) + 2sin^2(x) = 2

13. Substitute sin^2(2x) = 1 - cos^2(2x): 1 - cos^2(2x) + sin(4x) + 2sin^2(x) = 2

14. Rearrange the equation: 3sin^2(x) - cos^2(2x) + sin(4x) = 1

15. Use the identity cos^2(2x) = 1 - sin^2(2x): 3sin^2(x) + sin^2(2x) + sin(4x) = 1

16. Substitute sin(2x) = 2sin(x)cos(x): 3sin^2(x) + 2sin^2(x) + 2sin(x)cos(x)cos(2x) = 1

17. Substitute cos(2x) = 2cos^2(x) - 1: 3sin^2(x) + 2sin^2(x) + 2sin(x)cos(x)(2cos^2(x) - 1) = 1

18. Distribute and simplify: 3sin^2(x) + 2sin^2(x) + 4sin(x)cos^3(x) - 2sin(x)cos(x) = 1

19. Combine like terms: 5sin^2(x) + 2sin(x)cos(x) + 4sin(x)cos^3(x) = 1

At this point, the equation involves multiple trigonometric terms and doesn't seem to have a simple closed-form solution for x. Depending on the context or specific problem you're working on, there might be further simplifications or numerical methods that could be used to approximate solutions.