solve the following equation
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Answered by
5
Hi ,
Here I am using A instead of theta.
2Cos² A = 3sinA
2 ( 1 - sin² A ) = 3sinA
2 - 2sin²A = 3sinA
2sin²A + 3sinA - 2 = 0
2sin² A + 4sinA - sinA - 2 = 0
2sinA ( sinA + 2 ) - ( sinA + 2 ) = 0
( sinA + 2 ) ( 2sinA - 1 ) = 0
SinA + 2 = 0 or 2sinA - 1 = 0
sinA = - 2 or sinA = 1/2
SinA = sin 30°
A = 30°
I hope this helps you.
: )
Here I am using A instead of theta.
2Cos² A = 3sinA
2 ( 1 - sin² A ) = 3sinA
2 - 2sin²A = 3sinA
2sin²A + 3sinA - 2 = 0
2sin² A + 4sinA - sinA - 2 = 0
2sinA ( sinA + 2 ) - ( sinA + 2 ) = 0
( sinA + 2 ) ( 2sinA - 1 ) = 0
SinA + 2 = 0 or 2sinA - 1 = 0
sinA = - 2 or sinA = 1/2
SinA = sin 30°
A = 30°
I hope this helps you.
: )
Answered by
1
solution:
given: 2 cos²θ=3sinθ
cos²θ=3sinθ/2
2(1-sin²θ)=3sinθ
2-2sin²θ=3sinθ
2sin²θ+3sinθ-2=0
sinθ =
sinθ =
sinθ =
sinθ =
sinθ =
sinθ =
sinθ = sin30°
given: 2 cos²θ=3sinθ
cos²θ=3sinθ/2
2(1-sin²θ)=3sinθ
2-2sin²θ=3sinθ
2sin²θ+3sinθ-2=0
sinθ =
sinθ =
sinθ =
sinθ =
sinθ =
sinθ =
sinθ = sin30°
sivaprasath:
θ=30°
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