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

The general solution for 2sin^2(u)=1+sin(u),0<= y(u)<2pi is No Solution for u\in\mathbb{R}

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

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

No Solution for u \in R

Solution steps

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

Solve by substitution

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

sin (u) = 1, 0 \leq y (u) < 2 π :

No Solution

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

No Solution

Combine all the solutions

No Solution for u \in R

sin (α) = cos^{2} (4 5^{\circ})

0 = - 2 sin^{2} (x) + cos^{2} (x)

t, 3 cos (5 t) = 2

2 sin (x) cos (x) + 3 cos (x) = 0

tan (x) = 60

What is the general solution for 2sin^2(u)=1+sin(u),0<= y(u)<2pi ?
The general solution for 2sin^2(u)=1+sin(u),0<= y(u)<2pi is No Solution for u\in\mathbb{R}