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

The general solution for sqrt(2)sin^2(θ)-sin(θ)=0,0<= θ<2pi is θ= pi/4 ,θ=(3pi)/4 ,θ=0,θ=pi

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sqrt(2)sin^2(θ)-sin(θ)=0,0<= θ<2pi

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

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

Solution steps

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

Solve by substitution

sin (θ) = \frac{2}{2}, sin (θ) = 0

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

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

Combine all the solutions

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

cos (x^{2}) = \frac{1}{2}

sec^{2} (x) - 7 sec (x) = 0

e^{\frac{π}{2}} - e^{a r c t a n (1 + x^{2})} = 0

3 cot (θ) - 7 = 0

cos (x) = \frac{1249}{1250}

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