1)sin2x+2cosx=sinx+1 2)sinx*cosx-cos в квадрате x=1

Предыдущий
вопрос Следующий
вопрос

1) To solve the equation sin(2x) + 2cos(x) = sin(x) + 1:

We know that sin(2x) = 2sin(x)cos(x), so we can rewrite the equation as:

2sin(x)cos(x) + 2cos(x) = sin(x) + 1

Now, let's rewrite cos(x) as 1 - sin^2(x) using the Pythagorean identity.

2sin(x)(1 - sin^2(x)) + 2(1 - sin^2(x)) = sin(x) + 1

2sin(x) - 2sin^3(x) + 2 - 2sin^2(x) = sin(x) + 1

Rearranging the terms gives:

-2sin^3(x) + 2sin(x) - 2sin^2(x) + 2 = sin(x) + 1

Simplifying further:

-2sin^3(x) + 2sin(x) - 2sin^2(x) = 1

Now, we need to solve this cubic equation for sin(x).

2) To solve the equation sin(x)*cos(x) - cos^2(x) = 1:

Rewrite cos^2(x) as 1 - sin^2(x) using the Pythagorean identity:

sin(x)*cos(x) - (1 - sin^2(x)) = 1

Now simplify:

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

sin(x)*cos(x) + sin^2(x) = 2

Factor out sin(x):

sin(x)(cos(x) + sin(x)) = 2

Now, we need to solve this equation for sin(x).