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

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

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sin(2θ-pi/2)=-1,0<= θ<= 2pi

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

θ = π, θ = 2 π, θ = 0

Solution steps

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

General solutions for

sin (2 θ - \frac{π}{2}) = - 1

2 θ - \frac{π}{2} = \frac{3 π}{2} + 2 πn

Solve

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

θ = πn + π

Solutions for the range

0 \leq θ \leq 2 π

θ = π, θ = 2 π, θ = 0

sin (\frac{π}{11}) = sin (\frac{x π}{11})

3 + cos (4 θ) = 2 - cos (4 θ)

5 sec (θ) = 4 sec (θ) + 2

4 sec (π x) + 6 = - 2

x, y - 1 = - 4 + 3 sin (4 x - \frac{π}{4})

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