What is the general solution for sin(60)=cos(x+10) ?

The general solution for sin(60)=cos(x+10) is x=360n+20,x=360n+320

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sin(60)=cos(x+10)

Solution

sin (6 0^{\circ}) = cos (x + 1 0^{\circ})

Solution

x = 36 0^{\circ} n + 2 0^{\circ}, x = 36 0^{\circ} n + 32 0^{\circ}

Radians

x = \frac{π}{9} + 2 πn, x = \frac{16 π}{9} + 2 πn

Solution steps

sin (6 0^{\circ}) = cos (x + 1 0^{\circ})

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

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}

= \frac{3}{2}

= \frac{3}{2}

\frac{3}{2} = cos (x + 1 0^{\circ})

Switch sides

cos (x + 1 0^{\circ}) = \frac{3}{2}

General solutions for

cos (x + 1 0^{\circ}) = \frac{3}{2}

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

x + 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, x + 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n

x + 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, x + 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n

Solve

x + 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : x = 36 0^{\circ} n + 2 0^{\circ}

x + 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Move

1 0^{\circ}

to the right side

x + 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Subtract

1 0^{\circ}

from both sides

x + 1 0^{\circ} - 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

x + 1 0^{\circ} - 1 0^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

x + 1 0^{\circ} - 1 0^{\circ} : x

x + 1 0^{\circ} - 1 0^{\circ}

Add similar elements:

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

= x

Simplify

3 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 2 0^{\circ}

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

Group like terms

= 36 0^{\circ} n + 3 0^{\circ} - 1 0^{\circ}

Least Common Multiplier of

6, 18 : 18

6, 18

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

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

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

6

18

= 2 \cdot 3 \cdot 3

Multiply the numbers:

2 \cdot 3 \cdot 3 = 18

= 18

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

18

For

3 0^{\circ} :

multiply the denominator and numerator by

3

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

= 3 0^{\circ} - 1 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 - 18 0 ^{\circ}}{18}

Add similar elements:

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

= 2 0^{\circ}

Cancel the common factor:

2

= 36 0^{\circ} n + 2 0^{\circ}

x = 36 0^{\circ} n + 2 0^{\circ}

x = 36 0^{\circ} n + 2 0^{\circ}

x = 36 0^{\circ} n + 2 0^{\circ}

Solve

x + 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n : x = 36 0^{\circ} n + 32 0^{\circ}

x + 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n

Move

1 0^{\circ}

to the right side

x + 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n

Subtract

1 0^{\circ}

from both sides

x + 1 0^{\circ} - 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

x + 1 0^{\circ} - 1 0^{\circ} = 33 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

x + 1 0^{\circ} - 1 0^{\circ} : x

x + 1 0^{\circ} - 1 0^{\circ}

Add similar elements:

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

= x

Simplify

33 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 32 0^{\circ}

33 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Group like terms

= 36 0^{\circ} n - 1 0^{\circ} + 33 0^{\circ}

Least Common Multiplier of

18, 6 : 18

18, 6

Least Common Multiplier (LCM)

Prime factorization of

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

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

18

6

= 2 \cdot 3 \cdot 3

Multiply the numbers:

2 \cdot 3 \cdot 3 = 18

= 18

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

18

For

33 0^{\circ} :

multiply the denominator and numerator by

3

33 0^{\circ} = \frac{198 0 ^{\circ} 3}{6 \cdot 3} = 33 0^{\circ}

= - 1 0^{\circ} + 33 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} + 594 0 ^{\circ}}{18}

Add similar elements:

- 18 0^{\circ} + 594 0^{\circ} = 576 0^{\circ}

= 32 0^{\circ}

Cancel the common factor:

2

= 36 0^{\circ} n + 32 0^{\circ}

x = 36 0^{\circ} n + 32 0^{\circ}

x = 36 0^{\circ} n + 32 0^{\circ}

x = 36 0^{\circ} n + 32 0^{\circ}

x = 36 0^{\circ} n + 2 0^{\circ}, x = 36 0^{\circ} n + 32 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(60)=cos(x+10) ?
The general solution for sin(60)=cos(x+10) is x=360n+20,x=360n+320