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

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

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

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

Subtract

sin (x) + C

from both sides

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

Let:

u = \frac{x}{2}

2 sin^{2} (u) - sin (2 u) = 0

Rewrite using trig identities

2 sin^{2} (u) - 2 cos (u) sin (u) = 0

Factor

2 sin^{2} (u) - 2 cos (u) sin (u) : 2 sin (u) (sin (u) - cos (u))

2 sin (u) (sin (u) - cos (u)) = 0

Solving each part separately

sin (u) = 0 or sin (u) - cos (u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) - cos (u) = 0 : u = \frac{π}{4} + πn

Combine all the solutions

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

Substitute back

u = \frac{x}{2}

\frac{x}{2} = 2 πn : x = 4 πn

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

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

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

sin (a) = \frac{3}{5}, cos (a)

cos (x) = cos (4 x)

11 sin (B) + 13 = 4 sin (B) + 8

2 = 4 cos (x)

sin (2 x + π) = 0

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