Question

solve the following equation

Answer 1

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.

: )

Answer 2

solution:

given:  2 cos²θ=3sinθ

cos²θ=3sinθ/2
2(1-sin²θ)=3sinθ
2-2sin²θ=3sinθ
2sin²θ+3sinθ-2=0

sinθ = $\frac{-(3)+ \sqrt{(3)^2-4(2)(-2)} }{2(2)}$

sinθ = $\frac{-3+ \sqrt{9+16} }{4}$

sinθ = $\frac{-3+ \sqrt{25} }{4}$
sinθ = $\frac{-3+5}{4}$
sinθ = $\frac{2}{4}$
sinθ = $\frac{1}{2}$

sinθ = sin30°