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= pi/6 ,θ=(5pi)/6 ,θ=0,θ=pi

Popular >

2sin^2(θ)-sin(θ)=0,0<= θ<2pi

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

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = 0, θ = π

Solution steps

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

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 0

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

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

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = 0, θ = π

5 sin^{2} (x) + 2 sin (x) = 0

x, y^{3} + 7 y^{2} + y + 7 = 2 cos (x)

cos (2 x) - 11 cos (x) + 6 = 0

x, sin (x) = \frac{1}{2}

- 2 cos (x) + 1 = 5 cos (x) + 0

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= pi/6 ,θ=(5pi)/6 ,θ=0,θ=pi