Question

Solve. Between [0, 2pi] 2cos^2x+7cosx+3=0

Answer 1

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Answer:

  x = {2π/3, 4π/3}

Step-by-step explanation:

The equation can be factored as ...

  (cos(x) +3)(2cos(x) +1) = 0

The solutions are the values of x where the factors are zero.

  cos(x) +3 = 0

  cos(x) = -3 . . . . . . no real solution

__

  2cos(x) +1 = 0

  cos(x) = -1/2

  x = arccos(-1/2) = 2π/3

The other solution in the desired interval is ...

  x = 2π -(2π/3) = 4π/3

The solutions are x∈ {2π/3, 4π/3}.