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

The general solution for 2sin^2(θ)=sin(θ),0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=0,θ=pi,θ=2pi

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2sin^2(θ)=sin(θ),0<= θ<= 2pi

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

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = 0, θ = π, θ = 2 π

Solution steps

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

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 0

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

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

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = 0, θ = π, θ = 2 π

cos (2 x) = cos (4 x)

14 sin^{2} (x) - 7 sin (x) = 0

cos (α) = \frac{2}{3}

sec (- 2 x) = - 2

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

What is the general solution for 2sin^2(θ)=sin(θ),0<= θ<= 2pi ?
The general solution for 2sin^2(θ)=sin(θ),0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=0,θ=pi,θ=2pi