What is the general solution for 2sin^2(θ)-3sin(θ)+1=0,0<= θ<= 2pi ?

The general solution for 2sin^2(θ)-3sin(θ)+1=0,0<= θ<= 2pi is θ= pi/2 ,θ= pi/6 ,θ=(5pi)/6

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2sin^2(θ)-3sin(θ)+1=0,0<= θ<= 2pi

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

Solution steps

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

Solve by substitution

sin (θ) = 1, sin (θ) = \frac{1}{2}

sin (θ) = 1, 0 \leq θ \leq 2 π : θ = \frac{π}{2}

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{5 π}{6}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

2 sec (x) + 2 tan (x) = 2

2 sin^{2} (θ) + 3 sin (θ) = 0

cos^{2} (x) = sin^{2} (\frac{x}{2})

sin (x + \frac{π}{4}) = - \frac{2}{2}

1 = tan (θ)

What is the general solution for 2sin^2(θ)-3sin(θ)+1=0,0<= θ<= 2pi ?
The general solution for 2sin^2(θ)-3sin(θ)+1=0,0<= θ<= 2pi is θ= pi/2 ,θ= pi/6 ,θ=(5pi)/6