Solve for theta: 2sin^2theta-2cos theta=1/2
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Given trigonometric equation :
2 Sin² Ф - 2 Cos Ф = 1/2
2 ( 1- cos²Ф) - 2 cosФ = 1/2
2 Cos² Ф + 2 Cos Ф - 3/2 = 0
4 Cos² Ф + 4 Cos Ф - 3 = 0
Cos Ф = [-2 + √(4+12) ] /4
= 1/2 or -3/2
Since CosФ is always between -1 and 1, : Cos Ф = 1/2
Ф = 60°
2 Sin² Ф - 2 Cos Ф = 1/2
2 ( 1- cos²Ф) - 2 cosФ = 1/2
2 Cos² Ф + 2 Cos Ф - 3/2 = 0
4 Cos² Ф + 4 Cos Ф - 3 = 0
Cos Ф = [-2 + √(4+12) ] /4
= 1/2 or -3/2
Since CosФ is always between -1 and 1, : Cos Ф = 1/2
Ф = 60°
kvnmurty:
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