What is the general solution for 2sin^2(u)=1+sin(u),0<= t<2pi ?

The general solution for 2sin^2(u)=1+sin(u),0<= t<2pi is u= pi/2 ,u=(7pi)/6 ,u=(11pi)/6

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2sin^2(u)=1+sin(u),0<= t<2pi

2 sin^{2} (u) = 1 + sin (u), 0 \leq t < 2 π

u = \frac{π}{2}, u = \frac{7 π}{6}, u = \frac{11 π}{6}

Solution steps

2 sin^{2} (u) = 1 + sin (u), 0 \leq t < 2 π

Solve by substitution

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

sin (u) = 1, 0 \leq t < 2 π : u = \frac{π}{2}

sin (u) = - \frac{1}{2}, 0 \leq t < 2 π : u = \frac{7 π}{6}, u = \frac{11 π}{6}

Combine all the solutions

u = \frac{π}{2}, u = \frac{7 π}{6}, u = \frac{11 π}{6}

tan (3 y + 60) tan (2 y + 5) = 1

1 8^{2} = 8^{2} + 2 0^{2} - 2 \cdot 8 \cdot 20 \cdot cos (x)

cot (θ) = - \frac{π}{6}

3 tan^{2} (x) - 3 tan (x) = 0

2 sin^{2} (x) + 2 sin (x) + 1 = 0

What is the general solution for 2sin^2(u)=1+sin(u),0<= t<2pi ?
The general solution for 2sin^2(u)=1+sin(u),0<= t<2pi is u= pi/2 ,u=(7pi)/6 ,u=(11pi)/6