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

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

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2sin^2(θ)+sin(θ)-1=0,0<= θ<= 2pi

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

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{3 π}{2}

Solution steps

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

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{3 π}{2}

10 sin (x) cos (x) - 7 cos (x) = 0

sin (x + \frac{π}{4}) - sin (x - \frac{π}{4}) = 1

2 cos^{3} (x) = 2 cos (x)

sin^{2} (x) + cos^{2} (x) = 2

cot (\frac{2 θ}{3}) = - 3

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