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=0

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sin(2θ-pi/2)=-1,0<= θ<2pi

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

θ = π, θ = 0

Solution steps

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

θ = π, θ = 0

18 cos (x) - 93 = 0

2 sin (3 θ) - 1 = 0

cot^{2} (θ) + 1 = \frac{3}{4}

- 2 cos (x) - cos (2 x) \cdot 2 = 0

tan^{2} (x) + 4 tan (x) + 2 = 0

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=0