What is the general solution for 2*sin(2x+15)=1 ?

The general solution for 2*sin(2x+15)=1 is x=180n+7.5,x=180n+67.5

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2*sin(2x+15)=1

Solution

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

Solution

x = 18 0^{\circ} n + 7. 5^{\circ}, x = 18 0^{\circ} n + 67. 5^{\circ}

Radians

x = \frac{π}{24} + πn, x = \frac{3 π}{8} + πn

Solution steps

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (2 x + 1 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 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Solve

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 7. 5^{\circ}

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Move

1 5^{\circ}

to the right side

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

Subtract

1 5^{\circ}

from both sides

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

Simplify

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

Simplify

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

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

Add similar elements:

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

= 2 x

Simplify

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

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

Group like terms

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

Least Common Multiplier of

6, 12 : 12

6, 12

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

12 : 2 \cdot 2 \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

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

6

12

= 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}

= 3 0^{\circ} - 1 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}}{12}

Add similar elements:

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

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

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

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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}

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

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

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

Solve

2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 67. 5^{\circ}

2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Move

1 5^{\circ}

to the right side

2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Subtract

1 5^{\circ}

from both sides

2 x + 1 5^{\circ} - 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 1 5^{\circ}

Simplify

2 x + 1 5^{\circ} - 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 1 5^{\circ}

Simplify

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

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

Add similar elements:

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

= 2 x

Simplify

15 0^{\circ} + 36 0^{\circ} n - 1 5^{\circ} : 36 0^{\circ} n + 13 5^{\circ}

15 0^{\circ} + 36 0^{\circ} n - 1 5^{\circ}

Group like terms

= 36 0^{\circ} n + 15 0^{\circ} - 1 5^{\circ}

Least Common Multiplier of

6, 12 : 12

6, 12

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

12 : 2 \cdot 2 \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

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

6

12

= 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

15 0^{\circ} :

multiply the denominator and numerator by

2

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

= 15 0^{\circ} - 1 5^{\circ}

Since the denominators are equal, combine the fractions:

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

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

Add similar elements:

180 0^{\circ} - 18 0^{\circ} = 162 0^{\circ}

= 13 5^{\circ}

Cancel the common factor:

3

= 36 0^{\circ} n + 13 5^{\circ}

2 x = 36 0^{\circ} n + 13 5^{\circ}

2 x = 36 0^{\circ} n + 13 5^{\circ}

2 x = 36 0^{\circ} n + 13 5^{\circ}

Divide both sides by

2

2 x = 36 0^{\circ} n + 13 5^{\circ}

Divide both sides by

2

\frac{2 x}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{13 5 ^{\circ}}{2}

Simplify

\frac{2 x}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{13 5 ^{\circ}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{36 0 ^{\circ} n}{2} + \frac{13 5 ^{\circ}}{2} : 18 0^{\circ} n + 67. 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{13 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{13 5 ^{\circ}}{2} = 67. 5^{\circ}

\frac{13 5 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= 67. 5^{\circ}

= 18 0^{\circ} n + 67. 5^{\circ}

x = 18 0^{\circ} n + 67. 5^{\circ}

x = 18 0^{\circ} n + 67. 5^{\circ}

x = 18 0^{\circ} n + 67. 5^{\circ}

x = 18 0^{\circ} n + 7. 5^{\circ}, x = 18 0^{\circ} n + 67. 5^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 2*sin(2x+15)=1 ?
The general solution for 2*sin(2x+15)=1 is x=180n+7.5,x=180n+67.5