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Popular >

2sin^2(θ)+sin(θ)-1=0,0<= θ<2pi

Solution

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

Solution

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

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Solution steps

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

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

sin (θ) = - 1, 0 \leq θ < 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}

Popular Examples

2cos^2(x)-3cos(x)=2

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

cot(θ)= 1/(sqrt(3))

cot (θ) = \frac{1}{3}

-3sin^2(θ)+4sin(θ)-7=-6

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

sin(2x)-cos(2x)=1

sin (2 x) - cos (2 x) = 1

sin(x+pi/6)+cos(x+pi/6)=1

sin (x + \frac{π}{6}) + cos (x + \frac{π}{6}) = 1

Frequently Asked Questions (FAQ)

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