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

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

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sin(θ/2)-cos(θ)=0

sin (\frac{θ}{2}) - cos (θ) = 0

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

Solution steps

sin (\frac{θ}{2}) - cos (θ) = 0

Let:

u = \frac{θ}{2}

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

Rewrite using trig identities

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

Solve by substitution

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

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

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

\frac{θ}{2} = \frac{3 π}{2} + 2 πn : θ = 3 π + 4 πn

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

12 sec^{2} (x) - 16 = 0

cos (3 x - \frac{π}{8}) = \frac{2}{2}

5 (1 - cos (θ)) = sin^{2} (θ)

tan (x) - sec (x) = 1

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

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