Question

2cos^2x+3sinx=2 срочно надо

Answer 1

2cos²x+3sinx=2
2(1-sin²x)+3sinx-2=0
2-2sin²x+3sinx-2=0
3sinx-2sin²x=0
sinx(3-2sinx)=0
sinx=0  x=πn, n∈Z
3-2sinx=0  2sinx=3  sinx=3:2  sinx=1,5>1 чего быть не может
ответ: х=πn, n∈Z

Answer 2

Solving the Equation 2cos^2x + 3sinx = 2

To solve the equation 2cos^2x + 3sinx = 2, we can use trigonometric identities and equations to simplify and find the solution.

First, let's manipulate the equation using the identity cos^2x + sin^2x = 1: 2cos^2x + 3sinx = 2 2(1 - sin^2x) + 3sinx = 2 2 - 2sin^2x + 3sinx = 2 -2sin^2x + 3sinx = 0 sinx(3 - 2sinx) = 0

Now, we have two possibilities: 1. sinx = 0 2. 3 - 2sinx = 0

Let's solve for x in each case.

1. When sinx = 0: - This occurs when x = nπ, where n is an integer.

2. When 3 - 2sinx = 0: - Solving for sinx, we get sinx = 3/2, which is not a valid solution since the range of sinx is -1 to 1.

Therefore, the solutions to the equation 2cos^2x + 3sinx = 2 are x = nπ, where n is an integer.

This solution is based on trigonometric identities and equations.

I hope this helps! If you have any further questions, feel free to ask.