What is the general solution for sin(x)=cos(2x+51) ?

The general solution for sin(x)=cos(2x+51) is x=(21600n+2340)/(180),x=-(8460+21600n)/(60)

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sin(x)=cos(2x+51)

Solution

sin (x) = cos (2 x + 5 1^{\circ})

Solution

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}, x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

Radians

x = \frac{\frac{13 π}{5}}{36} + \frac{120 π}{180} n, x = - \frac{\frac{47 π}{5}}{12} - \frac{120 π}{60} n

Solution steps

sin (x) = cos (2 x + 5 1^{\circ})

Rewrite using trig identities

sin (x) = cos (2 x + 5 1^{\circ})

Use the following identity:

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

sin (x) = sin (9 0^{\circ} - (2 x + 5 1^{\circ}))

sin (x) = sin (9 0^{\circ} - (2 x + 5 1^{\circ}))

Apply trig inverse properties

sin (x) = sin (9 0^{\circ} - (2 x + 5 1^{\circ}))

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

x = 9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n, x = 18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n

x = 9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n, x = 18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n

x = 9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n : x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

x = 9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n

Expand

9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n : - 2 x + 36 0^{\circ} n + 3 9^{\circ}

9 0^{\circ} - (2 x + 5 1^{\circ}) + 36 0^{\circ} n

- (2 x + 5 1^{\circ}) : - 2 x - 5 1^{\circ}

- (2 x + 5 1^{\circ})

Distribute parentheses

= - (2 x) - (5 1^{\circ})

Apply minus-plus rules

+ (- a) = - a

= - 2 x - 5 1^{\circ}

= 9 0^{\circ} - 2 x - 5 1^{\circ} + 36 0^{\circ} n

Simplify

9 0^{\circ} - 2 x - 5 1^{\circ} + 36 0^{\circ} n : - 2 x + 36 0^{\circ} n + 3 9^{\circ}

9 0^{\circ} - 2 x - 5 1^{\circ} + 36 0^{\circ} n

Group like terms

= - 2 x + 36 0^{\circ} n + 9 0^{\circ} - 5 1^{\circ}

Least Common Multiplier of

2, 60 : 60

2, 60

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

60 : 2 \cdot 2 \cdot 3 \cdot 5

60

60

divides by

260 = 30 \cdot 2

= 2 \cdot 30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 5

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

2

60

= 2 \cdot 2 \cdot 3 \cdot 5

Multiply the numbers:

2 \cdot 2 \cdot 3 \cdot 5 = 60

= 60

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

60

For

9 0^{\circ} :

multiply the denominator and numerator by

30

9 0^{\circ} = \frac{18 0 ^{\circ} 30}{2 \cdot 30} = 9 0^{\circ}

= 9 0^{\circ} - 5 1^{\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} 30 - 306 0 ^{\circ}}{60}

Add similar elements:

540 0^{\circ} - 306 0^{\circ} = 234 0^{\circ}

= - 2 x + 36 0^{\circ} n + 3 9^{\circ}

= - 2 x + 36 0^{\circ} n + 3 9^{\circ}

x = - 2 x + 36 0^{\circ} n + 3 9^{\circ}

Move

2 x

to the left side

x = - 2 x + 36 0^{\circ} n + 3 9^{\circ}

Add

2 x

to both sides

x + 2 x = - 2 x + 36 0^{\circ} n + 3 9^{\circ} + 2 x

Simplify

3 x = 36 0^{\circ} n + 3 9^{\circ}

3 x = 36 0^{\circ} n + 3 9^{\circ}

Divide both sides by

3

3 x = 36 0^{\circ} n + 3 9^{\circ}

Divide both sides by

3

\frac{3 x}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{3 9 ^{\circ}}{3}

Simplify

\frac{3 x}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{3 9 ^{\circ}}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{36 0 ^{\circ} n}{3} + \frac{3 9 ^{\circ}}{3} : \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

\frac{36 0 ^{\circ} n}{3} + \frac{3 9 ^{\circ}}{3}

Apply rule

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

= \frac{36 0 ^{\circ} n + 3 9 ^{\circ}}{3}

Join

36 0^{\circ} n + 3 9^{\circ} : \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{60}

36 0^{\circ} n + 3 9^{\circ}

Convert element to fraction:

36 0^{\circ} n = \frac{36 0 ^{\circ} n 60}{60}

= \frac{36 0 ^{\circ} n \cdot 60}{60} + 3 9^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{36 0 ^{\circ} n \cdot 60 + 234 0 ^{\circ}}{60}

Multiply the numbers:

2 \cdot 60 = 120

= \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{60}

= \frac{\frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{60}}{3}

Apply the fraction rule:

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

= \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{60 \cdot 3}

Multiply the numbers:

60 \cdot 3 = 180

= \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}

x = 18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n : x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = 18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n

Expand

18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n : 18 0^{\circ} + 2 x - 3 9^{\circ} + 36 0^{\circ} n

18 0^{\circ} - (9 0^{\circ} - (2 x + 5 1^{\circ})) + 36 0^{\circ} n

Expand

9 0^{\circ} - (2 x + 5 1^{\circ}) : - 2 x + 3 9^{\circ}

9 0^{\circ} - (2 x + 5 1^{\circ})

- (2 x + 5 1^{\circ}) : - 2 x - 5 1^{\circ}

- (2 x + 5 1^{\circ})

Distribute parentheses

= - (2 x) - (5 1^{\circ})

Apply minus-plus rules

+ (- a) = - a

= - 2 x - 5 1^{\circ}

= 9 0^{\circ} - 2 x - 5 1^{\circ}

Simplify

9 0^{\circ} - 2 x - 5 1^{\circ} : - 2 x + 3 9^{\circ}

9 0^{\circ} - 2 x - 5 1^{\circ}

Group like terms

= - 2 x + 9 0^{\circ} - 5 1^{\circ}

Least Common Multiplier of

2, 60 : 60

2, 60

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

60 : 2 \cdot 2 \cdot 3 \cdot 5

60

60

divides by

260 = 30 \cdot 2

= 2 \cdot 30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 5

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

2

60

= 2 \cdot 2 \cdot 3 \cdot 5

Multiply the numbers:

2 \cdot 2 \cdot 3 \cdot 5 = 60

= 60

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

60

For

9 0^{\circ} :

multiply the denominator and numerator by

30

9 0^{\circ} = \frac{18 0 ^{\circ} 30}{2 \cdot 30} = 9 0^{\circ}

= 9 0^{\circ} - 5 1^{\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} 30 - 306 0 ^{\circ}}{60}

Add similar elements:

540 0^{\circ} - 306 0^{\circ} = 234 0^{\circ}

= - 2 x + 3 9^{\circ}

= - 2 x + 3 9^{\circ}

= 18 0^{\circ} - (- 2 x + 3 9^{\circ}) + 36 0^{\circ} n

- (- 2 x + 3 9^{\circ}) : 2 x - 3 9^{\circ}

- (- 2 x + 3 9^{\circ})

Distribute parentheses

= - (- 2 x) - (3 9^{\circ})

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 2 x - 3 9^{\circ}

= 18 0^{\circ} + 2 x - 3 9^{\circ} + 36 0^{\circ} n

x = 18 0^{\circ} + 2 x - 3 9^{\circ} + 36 0^{\circ} n

Move

2 x

to the left side

x = 18 0^{\circ} + 2 x - 3 9^{\circ} + 36 0^{\circ} n

Subtract

2 x

from both sides

x - 2 x = 18 0^{\circ} + 2 x - 3 9^{\circ} + 36 0^{\circ} n - 2 x

Simplify

- x = 18 0^{\circ} - 3 9^{\circ} + 36 0^{\circ} n

- x = 18 0^{\circ} - 3 9^{\circ} + 36 0^{\circ} n

Divide both sides by

- 1

- x = 18 0^{\circ} - 3 9^{\circ} + 36 0^{\circ} n

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{18 0 ^{\circ}}{- 1} - \frac{3 9 ^{\circ}}{- 1} + \frac{36 0 ^{\circ} n}{- 1}

Simplify

\frac{- x}{- 1} = \frac{18 0 ^{\circ}}{- 1} - \frac{3 9 ^{\circ}}{- 1} + \frac{36 0 ^{\circ} n}{- 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{18 0 ^{\circ}}{- 1} - \frac{3 9 ^{\circ}}{- 1} + \frac{36 0 ^{\circ} n}{- 1} : - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

\frac{18 0 ^{\circ}}{- 1} - \frac{3 9 ^{\circ}}{- 1} + \frac{36 0 ^{\circ} n}{- 1}

Apply rule

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

= \frac{18 0 ^{\circ} - 3 9 ^{\circ} + 36 0 ^{\circ} n}{- 1}

Apply the fraction rule:

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

= - \frac{18 0 ^{\circ} - 3 9 ^{\circ} + 36 0 ^{\circ} n}{1}

Join

18 0^{\circ} - 3 9^{\circ} + 36 0^{\circ} n : \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

18 0^{\circ} - 3 9^{\circ} + 36 0^{\circ} n

Convert element to fraction:

18 0^{\circ} = 18 0^{\circ}, 36 0^{\circ} n = \frac{36 0 ^{\circ} n 60}{60}

= 18 0^{\circ} - 3 9^{\circ} + \frac{36 0 ^{\circ} n \cdot 60}{60}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 60 - 234 0 ^{\circ} + 36 0 ^{\circ} n \cdot 60}{60}

18 0^{\circ} 60 - 234 0^{\circ} + 36 0^{\circ} n \cdot 60 = 846 0^{\circ} + 2160 0^{\circ} n

18 0^{\circ} 60 - 234 0^{\circ} + 36 0^{\circ} n \cdot 60

Add similar elements:

1080 0^{\circ} - 234 0^{\circ} = 846 0^{\circ}

= 846 0^{\circ} + 2 \cdot 1080 0^{\circ} n

Multiply the numbers:

2 \cdot 60 = 120

= 846 0^{\circ} + 2160 0^{\circ} n

= \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

= - \frac{\frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}, x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

x = \frac{2160 0 ^{\circ} n + 234 0 ^{\circ}}{180}, x = - \frac{846 0 ^{\circ} + 2160 0 ^{\circ} n}{60}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)=cos(2x+51) ?
The general solution for sin(x)=cos(2x+51) is x=(21600n+2340)/(180),x=-(8460+21600n)/(60)