Question

2cos2x+cos x-1=0 gt3x=1

Answer 1

Объяснение:

ну как???????????????

Answer 2

It looks like you have two separate equations. Let's solve each of them step by step:

1. Equation 1: 2cos^2(x) + cos(x) - 1 = 0

To solve this quadratic equation, we can use factoring. It's in the form of ax^2 + bx + c = 0, where a = 2, b = 1, and c = -1.

The equation can be factored as follows:

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

Now, set each factor to zero and solve for cos(x):

1. 2cos(x) - 1 = 0 2cos(x) = 1 cos(x) = 1/2

2. cos(x) + 1 = 0 cos(x) = -1

So, the solutions for cos(x) are cos(x) = 1/2 and cos(x) = -1.

2. Equation 2: gt3x = 1

I'm not sure what "gt3x" means in this context, as it's not a standard mathematical notation. If you meant "3x," then to solve for x, we can proceed as follows:

3x = 1 x = 1/3

So the solution for the second equation is x = 1/3.

Please clarify the "gt3x" part if it's not a typo or provide additional information if you have another equation or specific instructions.