What is the general solution for sin(3θ+72)=cos(48) ?

The general solution for sin(3θ+72)=cos(48) is θ=(10800n+1260-2160)/(90),θ=(4140+10800n-2160)/(90)

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Popular Trigonometry >

sin(3θ+72)=cos(48)

Solution

sin (3 θ + 72) = cos (4 8^{\circ})

Solution

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}, θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

Radians

θ = \frac{\frac{7 π}{5}}{18} - 24 + \frac{60 π}{90} n, θ = - 24 + \frac{\frac{23 π}{5}}{18} + \frac{60 π}{90} n

Solution steps

sin (3 θ + 72) = cos (4 8^{\circ})

Rewrite using trig identities

cos (4 8^{\circ})

Use the following identity:

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

sin (9 0^{\circ} - 4 8^{\circ})

sin (3 θ + 72) = sin (9 0^{\circ} - 4 8^{\circ})

Apply trig inverse properties

sin (3 θ + 72) = sin (9 0^{\circ} - 4 8^{\circ})

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

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n, 3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n, 3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n : θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n

Move

72

to the right side

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n

Subtract

72

from both sides

3 θ + 72 - 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n - 72

Simplify

3 θ + 72 - 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n - 72

Simplify

3 θ + 72 - 72 : 3 θ

3 θ + 72 - 72

Add similar elements:

72 - 72 = 0

= 3 θ

Simplify

9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n - 72 : 36 0^{\circ} n + 4 2^{\circ} - 72

9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n - 72

Combine the fractions

9 0^{\circ} - 4 8^{\circ} : 4 2^{\circ}

9 0^{\circ} - 4 8^{\circ}

Least Common Multiplier of

2, 15 : 30

2, 15

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

15 : 3 \cdot 5

15

15

divides by

315 = 5 \cdot 3

= 3 \cdot 5

3, 5

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 5

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

2

15

= 2 \cdot 3 \cdot 5

Multiply the numbers:

2 \cdot 3 \cdot 5 = 30

= 30

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

30

For

9 0^{\circ} :

multiply the denominator and numerator by

15

9 0^{\circ} = \frac{18 0 ^{\circ} 15}{2 \cdot 15} = 9 0^{\circ}

For

4 8^{\circ} :

multiply the denominator and numerator by

2

4 8^{\circ} = \frac{72 0 ^{\circ} 2}{15 \cdot 2} = 4 8^{\circ}

= 9 0^{\circ} - 4 8^{\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} 15 - 144 0 ^{\circ}}{30}

Add similar elements:

270 0^{\circ} - 144 0^{\circ} = 126 0^{\circ}

= 4 2^{\circ}

= 36 0^{\circ} n + 4 2^{\circ} - 72

3 θ = 36 0^{\circ} n + 4 2^{\circ} - 72

3 θ = 36 0^{\circ} n + 4 2^{\circ} - 72

3 θ = 36 0^{\circ} n + 4 2^{\circ} - 72

Divide both sides by

3

3 θ = 36 0^{\circ} n + 4 2^{\circ} - 72

Divide both sides by

3

\frac{3 θ}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{4 2 ^{\circ}}{3} - \frac{72}{3}

Simplify

\frac{3 θ}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{4 2 ^{\circ}}{3} - \frac{72}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{36 0 ^{\circ} n}{3} + \frac{4 2 ^{\circ}}{3} - \frac{72}{3} : \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

\frac{36 0 ^{\circ} n}{3} + \frac{4 2 ^{\circ}}{3} - \frac{72}{3}

Apply rule

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

= \frac{36 0 ^{\circ} n + 4 2 ^{\circ} - 72}{3}

Join

36 0^{\circ} n + 4 2^{\circ} - 72 : \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{30}

36 0^{\circ} n + 4 2^{\circ} - 72

Convert element to fraction:

36 0^{\circ} n = \frac{36 0 ^{\circ} n 30}{30}, 72 = \frac{72 \cdot 30}{30}

= \frac{36 0 ^{\circ} n \cdot 30}{30} + 4 2^{\circ} - \frac{72 \cdot 30}{30}

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 30 + 126 0 ^{\circ} - 72 \cdot 30}{30}

36 0^{\circ} n \cdot 30 + 126 0^{\circ} - 72 \cdot 30 = 1080 0^{\circ} n + 126 0^{\circ} - 2160

36 0^{\circ} n \cdot 30 + 126 0^{\circ} - 72 \cdot 30

Multiply the numbers:

2 \cdot 30 = 60

= 1080 0^{\circ} n + 126 0^{\circ} - 72 \cdot 30

Multiply the numbers:

72 \cdot 30 = 2160

= 1080 0^{\circ} n + 126 0^{\circ} - 2160

= \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{30}

= \frac{\frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{30}}{3}

Apply the fraction rule:

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

= \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{30 \cdot 3}

Multiply the numbers:

30 \cdot 3 = 90

= \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n : θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

Move

72

to the right side

3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

Subtract

72

from both sides

3 θ + 72 - 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n - 72

Simplify

3 θ + 72 - 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n - 72

Simplify

3 θ + 72 - 72 : 3 θ

3 θ + 72 - 72

Add similar elements:

72 - 72 = 0

= 3 θ

Simplify

18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n - 72 : 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n - 72

Join

9 0^{\circ} - 4 8^{\circ} : 4 2^{\circ}

9 0^{\circ} - 4 8^{\circ}

Least Common Multiplier of

2, 15 : 30

2, 15

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

15 : 3 \cdot 5

15

15

divides by

315 = 5 \cdot 3

= 3 \cdot 5

3, 5

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 5

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

2

15

= 2 \cdot 3 \cdot 5

Multiply the numbers:

2 \cdot 3 \cdot 5 = 30

= 30

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

30

For

9 0^{\circ} :

multiply the denominator and numerator by

15

9 0^{\circ} = \frac{18 0 ^{\circ} 15}{2 \cdot 15} = 9 0^{\circ}

For

4 8^{\circ} :

multiply the denominator and numerator by

2

4 8^{\circ} = \frac{72 0 ^{\circ} 2}{15 \cdot 2} = 4 8^{\circ}

= 9 0^{\circ} - 4 8^{\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} 15 - 144 0 ^{\circ}}{30}

Add similar elements:

270 0^{\circ} - 144 0^{\circ} = 126 0^{\circ}

= 4 2^{\circ}

= 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

3 θ = 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

3 θ = 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

3 θ = 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

Divide both sides by

3

3 θ = 18 0^{\circ} - 4 2^{\circ} + 36 0^{\circ} n - 72

Divide both sides by

3

\frac{3 θ}{3} = 6 0^{\circ} - \frac{4 2 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} - \frac{72}{3}

Simplify

\frac{3 θ}{3} = 6 0^{\circ} - \frac{4 2 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} - \frac{72}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

6 0^{\circ} - \frac{4 2 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} - \frac{72}{3} : \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

6 0^{\circ} - \frac{4 2 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} - \frac{72}{3}

Group like terms

= 6 0^{\circ} - \frac{72}{3} + \frac{36 0 ^{\circ} n}{3} - \frac{4 2 ^{\circ}}{3}

Apply rule

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

= \frac{18 0 ^{\circ} - 72 + 36 0 ^{\circ} n - 4 2 ^{\circ}}{3}

Join

18 0^{\circ} - 72 + 36 0^{\circ} n - 4 2^{\circ} : \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{30}

18 0^{\circ} - 72 + 36 0^{\circ} n - 4 2^{\circ}

Convert element to fraction:

18 0^{\circ} = 18 0^{\circ}, 72 = \frac{72 \cdot 30}{30}, 36 0^{\circ} n = \frac{36 0 ^{\circ} n 30}{30}

= 18 0^{\circ} - \frac{72 \cdot 30}{30} + \frac{36 0 ^{\circ} n \cdot 30}{30} - 4 2^{\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 - 72 \cdot 30 + 36 0 ^{\circ} n \cdot 30 - 126 0 ^{\circ}}{30}

18 0^{\circ} 30 - 72 \cdot 30 + 36 0^{\circ} n \cdot 30 - 126 0^{\circ} = 414 0^{\circ} + 1080 0^{\circ} n - 2160

18 0^{\circ} 30 - 72 \cdot 30 + 36 0^{\circ} n \cdot 30 - 126 0^{\circ}

Group like terms

= 540 0^{\circ} - 126 0^{\circ} + 2 \cdot 540 0^{\circ} n - 72 \cdot 30

Add similar elements:

540 0^{\circ} - 126 0^{\circ} = 414 0^{\circ}

= 414 0^{\circ} + 2 \cdot 540 0^{\circ} n - 72 \cdot 30

Multiply the numbers:

2 \cdot 30 = 60

= 414 0^{\circ} + 1080 0^{\circ} n - 72 \cdot 30

Multiply the numbers:

72 \cdot 30 = 2160

= 414 0^{\circ} + 1080 0^{\circ} n - 2160

= \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{30}

= \frac{\frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{30}}{3}

Apply the fraction rule:

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

= \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{30 \cdot 3}

Multiply the numbers:

30 \cdot 3 = 90

= \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}, θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(3θ+72)=cos(48) ?
The general solution for sin(3θ+72)=cos(48) is θ=(10800n+1260-2160)/(90),θ=(4140+10800n-2160)/(90)