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 \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Solution steps

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

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

sin^{2} (2 x) = 2 sin^{2} (x)

2 cos (x) + 2 sin (x) = 0

\frac{3}{2} = sin (arcsin (0) + c_{1})

tan^{2} (x) - sin^{2} (x) = tan^{2} (x) + sin^{2} (x)

3 cos (2 x) + cos (x) = 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