What is the general solution for 2sin^2(1/2 x)+C=2cos(x/2)sin(x/2)+C ?

The general solution for 2sin^2(1/2 x)+C=2cos(x/2)sin(x/2)+C is x=4pin,x=2pi+4pin,x= pi/2+2pin

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2sin^2(1/2 x)+C=2cos(x/2)sin(x/2)+C

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

x = 4 πn, x = 2 π + 4 πn, x = \frac{π}{2} + 2 πn

Solution steps

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

Subtract

2 cos (\frac{x}{2}) sin (\frac{x}{2}) + C

from both sides

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

Factor

2 sin^{2} (\frac{x}{2}) - 2 cos (\frac{x}{2}) sin (\frac{x}{2}) : 2 sin (\frac{x}{2}) (sin (\frac{x}{2}) - cos (\frac{x}{2}))

2 sin (\frac{x}{2}) (sin (\frac{x}{2}) - cos (\frac{x}{2})) = 0

Solving each part separately

sin (\frac{x}{2}) = 0 or sin (\frac{x}{2}) - cos (\frac{x}{2}) = 0

sin (\frac{x}{2}) = 0 : x = 4 πn, x = 2 π + 4 πn

sin (\frac{x}{2}) - cos (\frac{x}{2}) = 0 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = 4 πn, x = 2 π + 4 πn, x = \frac{π}{2} + 2 πn

sin (x) = 7

sin^{2} (x) cos^{2} (x) = \frac{2 - 2}{16}

(sin (x) - 1) (sin (x) - 5) = 0

(4 sin^{2} (x) - 1) (cos^{2} (x) - 1) = 0

3 sin^{2} (x) - 6 sin (x) = - 3, 0 \leq x \leq 2 π

What is the general solution for 2sin^2(1/2 x)+C=2cos(x/2)sin(x/2)+C ?
The general solution for 2sin^2(1/2 x)+C=2cos(x/2)sin(x/2)+C is x=4pin,x=2pi+4pin,x= pi/2+2pin