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Popular >

2sin^2(θ)+sqrt(3)sin(θ)=0,0<= θ<2pi

Solution

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

Solution

θ = 0, θ = π, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

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Solution steps

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

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{3}{2}

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

sin (θ) = - \frac{3}{2}, 0 \leq θ < 2 π : θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = 0, θ = π, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)+sqrt(3)sin(θ)=0,0<= θ<2pi ?
The general solution for 2sin^2(θ)+sqrt(3)sin(θ)=0,0<= θ<2pi is θ=0,θ=pi,θ=(4pi)/3 ,θ=(5pi)/3