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

The general solution for 2sin^2(θ)-3sin(θ)-2=0,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6

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2sin^2(θ)-3sin(θ)-2=0,0<= θ<2pi

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

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Solution steps

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

Solve by substitution

sin (θ) = 2, sin (θ) = - \frac{1}{2}

sin (θ) = 2, 0 \leq θ < 2 π :

No Solution

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) + cos (θ) = - 2

cot (2 x) = 1.732, 0 \leq x \leq 2 π

cot (θ) = 3.2404

8 sin^{2} (x) = 4

2 + 2 cos (2 x) = 0

What is the general solution for 2sin^2(θ)-3sin(θ)-2=0,0<= θ<2pi ?
The general solution for 2sin^2(θ)-3sin(θ)-2=0,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6