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

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

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

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

1 + sin (x)

from both sides

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

Rewrite using trig identities

- sin (x) = 0

Divide both sides by

- 1

sin (x) = 0

General solutions for

sin (x) = 0

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 2 πn, x = π + 2 πn

5 sin^{2} (x) - 6 sin (x) cos (x) = 0

10 sin (20 t - \frac{π}{6}) = 8

x, - cos (x) = 0

sin^{2} (x) tan^{2} (x) = tan^{2} (x) + sin^{2} (x)

θ, cot (θ) = - \frac{π}{6}

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