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

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

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sin(θ)-cos(θ/2)=0

sin (θ) - cos (\frac{θ}{2}) = 0

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

Solution steps

sin (θ) - cos (\frac{θ}{2}) = 0

Let:

u = \frac{θ}{2}

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

Rewrite using trig identities

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

Factor

- cos (u) + 2 cos (u) sin (u) : cos (u) (2 sin (u) - 1)

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

Solving each part separately

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

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

2 sin (u) - 1 = 0 : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

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

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

\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

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

tan (x) = \frac{195}{867}

x, cos (q x) = sin (r x)

sin (21 6^{\circ}) = sin (θ)

cos (θ) = - 0.2

sin^{2} (x) - 5 cos (x) = 3

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