What is the general solution for cos(2x-pi/2)=-1 ?

The general solution for cos(2x-pi/2)=-1 is x=pin+pi/2+pi/4

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cos(2x-pi/2)=-1

cos (2 x - \frac{π}{2}) = - 1

x = πn + \frac{π}{2} + \frac{π}{4}

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x - \frac{π}{2}) = - 1

General solutions for

cos (2 x - \frac{π}{2}) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x - \frac{π}{2} = π + 2 πn

2 x - \frac{π}{2} = π + 2 πn

Solve

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

2 x - \frac{π}{2} = π + 2 πn

Move

\frac{π}{2}

to the right side

2 x - \frac{π}{2} = π + 2 πn

Add

\frac{π}{2}

to both sides

2 x - \frac{π}{2} + \frac{π}{2} = π + 2 πn + \frac{π}{2}

Simplify

2 x = π + 2 πn + \frac{π}{2}

2 x = π + 2 πn + \frac{π}{2}

Divide both sides by

2

2 x = π + 2 πn + \frac{π}{2}

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

Simplify

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

\frac{π}{2} + \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

= \frac{π}{2} + πn + \frac{π}{4}

Group like terms

= πn + \frac{π}{2} + \frac{π}{4}

x = πn + \frac{π}{2} + \frac{π}{4}

x = πn + \frac{π}{2} + \frac{π}{4}

x = πn + \frac{π}{2} + \frac{π}{4}

x = πn + \frac{π}{2} + \frac{π}{4}

cot (x) = - \frac{1}{3}

4 cos^{2} (x) + 2 cos (x) - 2 = 0

tan (3 x) = 0

2 cos^{2} (θ) = 2 - cos (2 θ)

tan (x) = 2

What is the general solution for cos(2x-pi/2)=-1 ?
The general solution for cos(2x-pi/2)=-1 is x=pin+pi/2+pi/4