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=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2θ-pi/2)=-1,0<= θ<2pi

Solution

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

Solution

θ = π, θ = 0

Degrees

θ = 18 0^{\circ}, θ = 0^{\circ}

Solution steps

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

θ = π, θ = 0

Popular Examples

18cos(x)-9sqrt(3)=0

18 cos (x) - 93 = 0

sqrt(2)sin(3θ)-1=0

2 sin (3 θ) - 1 = 0

cot^2(θ)+1= 3/4

cot^{2} (θ) + 1 = \frac{3}{4}

-2cos(x)-cos(2x)*2=0

- 2 cos (x) - cos (2 x) \cdot 2 = 0

tan^2(x)+4tan(x)+2=0

tan^{2} (x) + 4 tan (x) + 2 = 0

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=0