Question

4cos^2x-3cosx-1=0 пожалуйста помогите ​

Answer 1

4cos²x-3cosx-1=0

примем косинус за некоторую переменную, допустим: cosx=a

4a²-3a-1=0

D=b²-4ac=9+4*4=25

a1;2= 3±5 / 2*4= 3±5 / 8= 1; -1/4

cosx=1

x=2πk, k∈Z

(частный случай)

или

cosx=-1/4

x = ±arccos(-1/4)+2πk

x = ± (π-arccos(1/4))+2πk, k∈Z

Answer 2

Solving the Equation 4cos^2x - 3cosx - 1 = 0

To solve the equation 4cos^2x - 3cosx - 1 = 0, we can use the quadratic formula or factorization. Let's solve it using the quadratic formula.

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the given equation 4cos^2x - 3cosx - 1 = 0, we have: a = 4, b = -3, and c = -1.

Using the quadratic formula: cosx = (-(-3) ± √((-3)^2 - 4*4*(-1))) / (2*4)

Solving for cosx gives us the solutions for the equation.

Solutions for cosx

The solutions for the equation 4cos^2x - 3cosx - 1 = 0 are: cosx = (3 ± √(25)) / 8

From the solutions, we can further simplify and find the exact values of cosx.

Conclusion

The solutions for the equation 4cos^2x - 3cosx - 1 = 0 are: cosx = (3 + √25) / 8 and cosx = (3 - √25) / 8

These are the solutions for the given equation.