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

The general solution for cos(2θ-pi/2)=-1,0<= θ<2pi is θ=(3pi)/4 ,θ=(7pi)/4

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cos(2θ-pi/2)=-1,0<= θ<2pi

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

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Solution steps

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

General solutions for

cos (2 θ - \frac{π}{2}) = - 1

2 θ - \frac{π}{2} = π + 2 πn

Solve

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

θ = πn + \frac{π}{2} + \frac{π}{4}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

sinh (x) = - \frac{3}{4}

tan (3 θ) = 1

7 - 7 sin (θ) = 7

- 12 cos^{2} (θ) - 8 = - 17

2 cos (\frac{1}{2} x) - 2 = 0

What is the general solution for cos(2θ-pi/2)=-1,0<= θ<2pi ?
The general solution for cos(2θ-pi/2)=-1,0<= θ<2pi is θ=(3pi)/4 ,θ=(7pi)/4