hannahbradley15
hannahbradley15 hannahbradley15
  • 10-07-2016
  • Mathematics
contestada

2 sin 2θ − 3 sin θ = 0

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needmathhelp11 needmathhelp11
  • 10-07-2016
(-1/4)  
.................................

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olemakpadu
olemakpadu olemakpadu
  • 10-07-2016
sin2θ = 2sinθcosθ

2 sin 2θ − 3 sin θ = 0

2(2sinθcosθ) − 3 sin θ = 0

4sinθcosθ - 3sinθ = 0

sinθ(4cosθ - 3) = 0

sinθ = 0,             4cosθ - 3 = 0    cosθ = 3/4

θ = sin⁻¹(0)            θ = cos⁻¹(0.75)


θ = 0, 180, 360 degrees,            θ ≈ 41.41°,   360 - 41.41 = 318.59°

θ = 0°, 180°, 360°, ≈ 41.41°, ≈318.59°
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