What is the general solution for sin(pi/2-x)=-1,0<= x<= 360 ?

Q: What is the general solution for sin(pi/2-x)=-1,0<= x<= 360 ?

The general solution for sin(pi/2-x)=-1,0<= x<= 360 is x=180

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sin(pi/2-x)=-1,0<= x<= 360

Solution

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 18 0^{\circ}

Radians

x = π

Solution steps

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

General solutions for

sin (9 0^{\circ} - x) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n

Solve

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n : x = - 36 0^{\circ} n - 18 0^{\circ}

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n

Move

9 0^{\circ}

to the right side

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n

Subtract

9 0^{\circ}

from both sides

9 0^{\circ} - x - 9 0^{\circ} = 27 0^{\circ} + 36 0^{\circ} n - 9 0^{\circ}

Simplify

9 0^{\circ} - x - 9 0^{\circ} = 27 0^{\circ} + 36 0^{\circ} n - 9 0^{\circ}

Simplify

9 0^{\circ} - x - 9 0^{\circ} : - x

9 0^{\circ} - x - 9 0^{\circ}

Add similar elements:

9 0^{\circ} - 9 0^{\circ} = 0

= - x

Simplify

27 0^{\circ} + 36 0^{\circ} n - 9 0^{\circ} : 36 0^{\circ} n + 18 0^{\circ}

27 0^{\circ} + 36 0^{\circ} n - 9 0^{\circ}

Group like terms

= 36 0^{\circ} n - 9 0^{\circ} + 27 0^{\circ}

Combine the fractions

- 9 0^{\circ} + 27 0^{\circ} : 18 0^{\circ}

Apply rule

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

= \frac{- 18 0 ^{\circ} + 54 0 ^{\circ}}{2}

Add similar elements:

- 18 0^{\circ} + 54 0^{\circ} = 36 0^{\circ}

= 18 0^{\circ}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ}

= 36 0^{\circ} n + 18 0^{\circ}

- x = 36 0^{\circ} n + 18 0^{\circ}

- x = 36 0^{\circ} n + 18 0^{\circ}

- x = 36 0^{\circ} n + 18 0^{\circ}

Divide both sides by

- 1

- x = 36 0^{\circ} n + 18 0^{\circ}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{36 0 ^{\circ} n}{- 1} + \frac{18 0 ^{\circ}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{36 0 ^{\circ} n}{- 1} + \frac{18 0 ^{\circ}}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{36 0 ^{\circ} n}{- 1} + \frac{18 0 ^{\circ}}{- 1} : - 36 0^{\circ} n - 18 0^{\circ}

\frac{36 0 ^{\circ} n}{- 1} + \frac{18 0 ^{\circ}}{- 1}

\frac{36 0 ^{\circ} n}{- 1} = - 36 0^{\circ} n

\frac{36 0 ^{\circ} n}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{36 0 ^{\circ} n}{1}

Apply rule

\frac{a}{1} = a

= - 36 0^{\circ} n

= - 36 0^{\circ} n + \frac{18 0 ^{\circ}}{- 1}

\frac{18 0 ^{\circ}}{- 1} = - 18 0^{\circ}

\frac{18 0 ^{\circ}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - 18 0^{\circ}

Apply rule

\frac{a}{1} = a

= - 18 0^{\circ}

= - 36 0^{\circ} n - 18 0^{\circ}

x = - 36 0^{\circ} n - 18 0^{\circ}

x = - 36 0^{\circ} n - 18 0^{\circ}

x = - 36 0^{\circ} n - 18 0^{\circ}

x = - 36 0^{\circ} n - 18 0^{\circ}

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 18 0^{\circ}

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6 (cos (b))^{2} + cos (b) - 1 = 0

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Frequently Asked Questions (FAQ)

What is the general solution for sin(pi/2-x)=-1,0<= x<= 360 ?
The general solution for sin(pi/2-x)=-1,0<= x<= 360 is x=180