1)6 cos^2x-sin^2x=5 2)sin^2x-2cos^2x+1/2sin2x=0 3)sin^4x-cos^4x=sin^2x 4)1-3cos^2x=2sinx cosx

Предыдущий
вопрос Следующий
вопрос

1) 6cos^2x - sin^2x = 5
We can rewrite sin^2x as 1 - cos^2x:
6cos^2x - (1 - cos^2x) = 5
6cos^2x - 1 + cos^2x = 5
7cos^2x - 1 = 5
7cos^2x = 6
cos^2x = 6/7
cosx = ±√(6/7)

2) sin^2x - 2cos^2x + 1/2sin2x = 0
Using the sin2x formula, sin2x = 2sinxcosx:
sin^2x - 2cos^2x + sinxcosx = 0

Unfortunately, this equation is not easily simplified into a straightforward solution.

3) sin^4x - cos^4x = sin^2x
We can rewrite sin^4x as (sin^2x)^2:
(sin^2x)^2 - cos^4x = sin^2x
(sin^2x - cos^2x)(sin^2x + cos^2x) = sin^2x
Using the identity sin^2x + cos^2x = 1:
sin^2x - cos^2x = sin^2x
cos^2x = 0

4) 1 - 3cos^2x = 2sinx*cosx
Using the Pythagorean identity sin^2x + cos^2x = 1:
1 - 3(1 - sin^2x) = 2sinx(√(1 - sin^2x))
1 - 3 + 3sin^2x = 2sinx(√(1 - sin^2x))
3sin^2x - 2sinx - 2 = 0

Unfortunately, this equation may not have a straightforward analytical solution and may require numerical methods to solve.