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

Q: 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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ-pi/2)=-1,0<= θ<= 2pi

Solution

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

Solution

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 31 5^{\circ}

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

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 θ - \frac{π}{2} = π + 2 πn

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}

2 θ - \frac{π}{2} = π + 2 πn

Move

\frac{π}{2}

to the right side

2 θ - \frac{π}{2} = π + 2 πn

Add

\frac{π}{2}

to both sides

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

Simplify

2 θ = π + 2 πn + \frac{π}{2}

2 θ = π + 2 πn + \frac{π}{2}

Divide both sides by

2

2 θ = π + 2 πn + \frac{π}{2}

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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}

θ = πn + \frac{π}{2} + \frac{π}{4}

θ = πn + \frac{π}{2} + \frac{π}{4}

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}

Popular Examples

sin^2(2x)=2sin^2(x)

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

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

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

(sqrt(3))/2 =sin(arcsin(0)+c_{1})

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

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

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

3cos(2x)+cos(x)=0

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

Frequently Asked Questions (FAQ)

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