Find all the solutions of the equation in the interval [0,2pi). 4sin^(2)x=5-4cosx

                       4sin²(x) = 5 - 4cos(x)
        4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
   4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
         4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
     4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
                2 - 2cos(2x) = 5 - 4cos(x)
              - 2                   - 2
                    -2cos(2x) = 3 - 4cos(x)
            -2[2cos²(x) - 1] = 3 - 4cos(x)
               -4cos²(x) + 2 = 3 - 4cos(x)
                               - 2  - 2
                     -4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
 4cos²(x) - 4cos(x) + 1 = 0
               [2cos(x) - 1]² = 0
                  2cos(x) - 1 = 0
                              + 1 + 1
                       2cos(x) = 1
                            2         2
                         cos(x) = ¹/₂
                cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
                                 x = 60, 300
                                 x = π/3, 5π/3

[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π

Your answer would be ahere is how: perimeter is how long all the sides are. in a rectangle, there are two sets of parallel sides. so, to find this, we would to the following.(15x2)+(4x2)30+838if you have questions ask! and mark me brainliest! i really wanna level up : 3