Maths (sin and cos)

Please read :

1.
Since 0 < t < π
Then 0 < 2t < 2π

cos(6t) - cos(2t) = sin(4t)
[4 cos³(2t) - 3 cos(2t)] - cos(2t) = 2 sin(2t) cos(2t)
4 cos³(2t) - 4 cos(2t) - 2 sin(2t) cos(2t) = 0
2 cos³(2t) - 2 cos(2t) - sin(2t) cos(2t) = 0
cos(2t) [2 cos²(2t) - 2 - sin(2t)] = 0
cos(2t) {2 [1 - sin²(2t)] - 2 - sin(2t)} = 0
cos(2t) {2 - 2 sin²(2t) - 2 - sin(2t)} = 0
-cos(2t) [2 sin²(2t) + sin(2t)] = 0
cos(2t) sin(2t) [2 sin(2t) + 1] = 0
cos(2t) = 0 or sin(2t) = 0 or sin(2t) = -1/2
2t = π/2, 3π/2 or 2t = π or 2t = 7π/6, 11π/6
t = π/4, π/2, 7π/12, 3π/4, 11π/12

(If 0 ≤ t ≤ π, t = 0, π/4, π/2, 7π/12,3π/4, 11π/12, π)


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2.
-2 sin(x) + 4 cos(x) = 2
-sin(x) + 2 cos(x) = 1
-2 sin(x/2) cos(x/2) + 2 [cos²(x/2) - sin²(x/2)]= cos²(x/2) + sin²(x/2)
-2 sin(x/2) cos(x/2) + 2 cos²(x/2) - 2 sin²(x/2)= cos²(x/2) + sin²(x/2)
3 sin²(x/2) + 2 sin(x/2) cos(x/2) - cos²(x/2) = 0
[3 sin²(x/2) + 2 sin(x/2) cos(x/2) - cos²(x/2)] / cos²(x/2)= 0
3tan²(x/2) + 2 tan(x/2) - 1 = 0
[3tan(x/2) - 1] [tan(x/2) + 1] = 0
tan(x/2) = 1/3 or tan(x/2) = -1

The first two positive solutions for x/2 :
x/2 = 0.322 (rad) or x/2 = 3π/4 (rad)

The first two positive solution for x :
x = 0.644 (rad) or x = 3π/2

0 < 2t <

cos(6t) - cos(2t) = sin(4t)
==> -2sin(4t)sin(2t) = sin(4t)
==> sin(4t) [2sin(2t) + 1] = 0
==> 4t = π, 2π, 3π or 2t = 7π/6, 11π/6
==> t = π/4, π/2, 3π/4, 7π/12, 11π/12