Question

2cos^2x+3cosx-2=0 решите пожалуйста, срочно, очень срочно нужно

Answer 1

2*cos²x+3*cosx-2=0               ОДЗ: cosx∈[-1;1]
Пусть cosx=t  
2t²+3t-2=0   D=25
 t₁=cosx=2 ∉ ОДЗ
t₂=cosx=-1/2  
x₁=2π/3+2πn         x₂=4π/3+2πn.

Answer 2

Solving the Equation 2cos^2x + 3cosx - 2 = 0

To solve the equation 2cos^2x + 3cosx - 2 = 0, we can use a substitution to simplify the equation and then solve for the variable.

First, let's use the substitution cosx = t to simplify the equation.

Substituting cosx = t, the equation becomes 2t^2 + 3t - 2 = 0.

Now, we can solve for t using the quadratic formula, which is given by:

t = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 2, b = 3, and c = -2.

Plugging these values into the quadratic formula, we get:

t = (-3 ± √(3^2 - 4*2*(-2))) / (2*2)

Solving further, we get:

t = (-3 ± √(9 + 16)) / 4

t = (-3 ± √25) / 4

This gives us two possible values for t:

t = (-3 + 5) / 4 and t = (-3 - 5) / 4

So, t = 1 and t = -2.

Therefore, the solutions for cosx are 1 and -2.

Conclusion

The solutions for the equation 2cos^2x + 3cosx - 2 = 0 are cosx = 1 and cosx = -2.