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Popular >

2sin^2(θ)+sin(θ)=0,0<= θ<2pi

Solution

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

Solution

θ = 0, θ = π, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

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Solution steps

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

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{1}{2}

sin (θ) = 0, 0 \leq θ < 2 π : θ = 0, θ = π

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 = 0, θ = π, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Popular Examples

sin(x)+cos(x)=0,-pi<= x<= pi

sin (x) + cos (x) = 0, - π \leq x \leq π

cos(θ)=(-1)/2 ,0<= θ<= 4pi

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

2 + 4 cos (θ) = 0

-cos(x)-4sin^2(x)+2=0

- cos (x) - 4 sin^{2} (x) + 2 = 0

2sin(t)+cos(2t)=0

2 sin (t) + cos (2 t) = 0

Frequently Asked Questions (FAQ)

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