What is the general solution for -8sin(2θ+45)=-4,0<= x<360 ?

Q: What is the general solution for -8sin(2θ+45)=-4,0<= x<360 ?

The general solution for -8sin(2θ+45)=-4,0<= x<360 is θ=52.5,θ=172.5,θ=232.5,θ=352.5

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Popular Trigonometry >

-8sin(2θ+45)=-4,0<= x<360

Solution

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

Solution

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

Radians

θ = \frac{7 π}{24}, θ = \frac{23 π}{24}, θ = \frac{31 π}{24}, θ = \frac{47 π}{24}

Solution steps

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

Divide both sides by

- 8

- 8 sin (2 θ + 4 5^{\circ}) = - 4

Divide both sides by

- 8

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} = \frac{- 4}{- 8}

Simplify

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} = \frac{- 4}{- 8}

Simplify

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} : sin (2 θ + 4 5^{\circ})

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8}

Apply the fraction rule:

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

= \frac{8 sin ( 2 θ + 4 5 ^{\circ} )}{8}

Divide the numbers:

\frac{8}{8} = 1

= sin (2 θ + 4 5^{\circ})

Simplify

\frac{- 4}{- 8} : \frac{1}{2}

\frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

General solutions for

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

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}

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Solve

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n - 7. 5^{\circ}

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Move

4 5^{\circ}

to the right side

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Subtract

4 5^{\circ}

from both sides

2 θ + 4 5^{\circ} - 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Simplify

2 θ + 4 5^{\circ} - 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Simplify

2 θ + 4 5^{\circ} - 4 5^{\circ} : 2 θ

2 θ + 4 5^{\circ} - 4 5^{\circ}

Add similar elements:

4 5^{\circ} - 4 5^{\circ} = 0

= 2 θ

Simplify

3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ} : 36 0^{\circ} n - 1 5^{\circ}

3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Group like terms

= 36 0^{\circ} n + 3 0^{\circ} - 4 5^{\circ}

Least Common Multiplier of

6, 4 : 12

6, 4

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

3 0^{\circ} :

multiply the denominator and numerator by

2

3 0^{\circ} = \frac{18 0 ^{\circ} 2}{6 \cdot 2} = 3 0^{\circ}

For

4 5^{\circ} :

multiply the denominator and numerator by

3

4 5^{\circ} = \frac{18 0 ^{\circ} 3}{4 \cdot 3} = 4 5^{\circ}

= 3 0^{\circ} - 4 5^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 2 - 18 0 ^{\circ} 3}{12}

Add similar elements:

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

= \frac{- 18 0 ^{\circ}}{12}

Apply the fraction rule:

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

= 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

Divide both sides by

2

2 θ = 36 0^{\circ} n - 1 5^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2} : 18 0^{\circ} n - 7. 5^{\circ}

\frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{1 5 ^{\circ}}{2} = 7. 5^{\circ}

\frac{1 5 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{12 \cdot 2}

Multiply the numbers:

12 \cdot 2 = 24

= 7. 5^{\circ}

= 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

Solve

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 52. 5^{\circ}

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Move

4 5^{\circ}

to the right side

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Subtract

4 5^{\circ}

from both sides

2 θ + 4 5^{\circ} - 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Simplify

2 θ + 4 5^{\circ} - 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Simplify

2 θ + 4 5^{\circ} - 4 5^{\circ} : 2 θ

2 θ + 4 5^{\circ} - 4 5^{\circ}

Add similar elements:

4 5^{\circ} - 4 5^{\circ} = 0

= 2 θ

Simplify

15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ} : 36 0^{\circ} n + 10 5^{\circ}

15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Group like terms

= 36 0^{\circ} n - 4 5^{\circ} + 15 0^{\circ}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

4

6

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

4 5^{\circ} :

multiply the denominator and numerator by

3

4 5^{\circ} = \frac{18 0 ^{\circ} 3}{4 \cdot 3} = 4 5^{\circ}

For

15 0^{\circ} :

multiply the denominator and numerator by

2

15 0^{\circ} = \frac{90 0 ^{\circ} 2}{6 \cdot 2} = 15 0^{\circ}

= - 4 5^{\circ} + 15 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{- 18 0 ^{\circ} 3 + 180 0 ^{\circ}}{12}

Add similar elements:

- 54 0^{\circ} + 180 0^{\circ} = 126 0^{\circ}

= 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

Divide both sides by

2

2 θ = 36 0^{\circ} n + 10 5^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2} : 18 0^{\circ} n + 52. 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{10 5 ^{\circ}}{2} = 52. 5^{\circ}

\frac{10 5 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{126 0 ^{\circ}}{12 \cdot 2}

Multiply the numbers:

12 \cdot 2 = 24

= 52. 5^{\circ}

= 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}, θ = 18 0^{\circ} n + 52. 5^{\circ}

Solutions for the range

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

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -8sin(2θ+45)=-4,0<= x<360 ?
The general solution for -8sin(2θ+45)=-4,0<= x<360 is θ=52.5,θ=172.5,θ=232.5,θ=352.5