What is the general solution for sin(θ/2)-1=0,0<= θ<= 2pi ?

The general solution for sin(θ/2)-1=0,0<= θ<= 2pi is θ=pi

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sin(θ/2)-1=0,0<= θ<= 2pi

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

θ = π

Solution steps

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

sin (\frac{θ}{2}) = 1

General solutions for

sin (\frac{θ}{2}) = 1

\frac{θ}{2} = \frac{π}{2} + 2 πn

Solve

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

θ = π + 4 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = π

cos (2 x) = - \frac{2}{2}, 0 \leq x \leq 2 π

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

cos (a) = 0

cos (3 x) = \frac{- 3}{2}

2 cos (2 θ) - 10 sin (θ) - 2 = - 6 sin (θ) - 3

What is the general solution for sin(θ/2)-1=0,0<= θ<= 2pi ?
The general solution for sin(θ/2)-1=0,0<= θ<= 2pi is θ=pi