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

Q: 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=2pi,θ=0

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sin(2θ-pi/2)=-1,0<= θ<= 2pi

Solution

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

Solution

θ = π, θ = 2 π, θ = 0

Degrees

θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 0^{\circ}

Solution steps

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

General solutions for

sin (2 θ - \frac{π}{2}) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ - \frac{π}{2} = \frac{3 π}{2} + 2 πn

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

2 θ - \frac{π}{2} = \frac{3 π}{2} + 2 πn

Move

\frac{π}{2}

to the right side

2 θ - \frac{π}{2} = \frac{3 π}{2} + 2 πn

Add

\frac{π}{2}

to both sides

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{3 π}{2} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{3 π}{2} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} : 2 θ

2 θ - \frac{π}{2} + \frac{π}{2}

Add similar elements:

- \frac{π}{2} + \frac{π}{2} = 0

= 2 θ

Simplify

\frac{3 π}{2} + 2 πn + \frac{π}{2} : 2 πn + 2 π

\frac{3 π}{2} + 2 πn + \frac{π}{2}

Group like terms

= 2 πn + \frac{π}{2} + \frac{3 π}{2}

Combine the fractions

\frac{π}{2} + \frac{3 π}{2} : 2 π

Apply rule

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

= \frac{π + 3 π}{2}

Add similar elements:

π + 3 π = 4 π

= \frac{4 π}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 π

= 2 πn + 2 π

2 θ = 2 πn + 2 π

2 θ = 2 πn + 2 π

2 θ = 2 πn + 2 π

Divide both sides by

2

2 θ = 2 πn + 2 π

Divide both sides by

2

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

Simplify

θ = πn + π

θ = πn + π

θ = πn + π

Solutions for the range

0 \leq θ \leq 2 π

θ = π, θ = 2 π, θ = 0

Popular Examples

sin(pi/(11))=sin((xpi)/(11))

sin (\frac{π}{11}) = sin (\frac{x π}{11})

3+cos(4θ)=2-cos(4θ)

3 + cos (4 θ) = 2 - cos (4 θ)

5sec(θ)=4sec(θ)+sqrt(2)

5 sec (θ) = 4 sec (θ) + 2

4sec(pix)+6=-2

4 sec (π x) + 6 = - 2

solvefor x,y-1=-4+3sin(4x-pi/4)solve for

x, y - 1 = - 4 + 3 sin (4 x - \frac{π}{4})

Frequently Asked Questions (FAQ)

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=2pi,θ=0