What is the general solution for 2sin^2(u)=1-sin(u) ?

The general solution for 2sin^2(u)=1-sin(u) is u= pi/6+2pin,u=(5pi)/6+2pin,u=(3pi)/2+2pin

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2sin^2(u)=1-sin(u)

2 sin^{2} (u) = 1 - sin (u)

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = \frac{3 π}{2} + 2 πn

Solution steps

2 sin^{2} (u) = 1 - sin (u)

Solve by substitution

sin (u) = \frac{1}{2}, sin (u) = - 1

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

Combine all the solutions

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = \frac{3 π}{2} + 2 πn

2 cos^{2} (x) = 1 - sin (x)

sin (t) = - \frac{3}{2}

2 sin (2 θ) - 3 sin (θ) = 0

cos^{2} (x) + cos^{2} (\frac{x}{2}) - \frac{1}{2} = 0

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

What is the general solution for 2sin^2(u)=1-sin(u) ?
The general solution for 2sin^2(u)=1-sin(u) is u= pi/6+2pin,u=(5pi)/6+2pin,u=(3pi)/2+2pin