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Maths (sin and cos)
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Q1, Find all solutions to cos(6t)-cos(2t) = sin(4t) on 0 < t < pi Q3 Solve -2sin(x)+4cos(x) = 2 for the first 2 positive solutions ?Thank you so much
最佳解答:
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) ==== 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
其他解答:
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/124E350C6F8B48ECA2
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