What is the general solution for sin(x/2)+cos(x)-1=0 ?

The general solution for sin(x/2)+cos(x)-1=0 is x=4pin,x=2pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin

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sin(x/2)+cos(x)-1=0

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

x = 4 πn, x = 2 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn

Solution steps

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

Let:

u = \frac{x}{2}

sin (u) + cos (2 u) - 1 = 0

Rewrite using trig identities

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

Solve by substitution

sin (u) = 0, sin (u) = \frac{1}{2}

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 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}{6} + 2 πn : x = \frac{π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{3} + 4 πn

x = 4 πn, x = 2 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn

3 sin (x) sec (x) = 2 sin (x)

y, x = sin (y)

2 sin (\frac{x}{2}) + 3 = 0

2 cos^{2} (x) + 3 sin (x) = 0

tan (3 x) - 1 = 0

What is the general solution for sin(x/2)+cos(x)-1=0 ?
The general solution for sin(x/2)+cos(x)-1=0 is x=4pin,x=2pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin