What is the general solution for sin(2x)=cos(x-15) ?

The general solution for sin(2x)=cos(x-15) is x=(4320n+1260)/(36),x=180-105+360n

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sin(2x)=cos(x-15)

Solution

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

Solution

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}, x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

Radians

x = \frac{7 π}{36} + \frac{24 π}{36} n, x = π - \frac{7 π}{12} + 2 πn

Solution steps

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

Rewrite using trig identities

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

Use the following identity:

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

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

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

Apply trig inverse properties

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

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

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

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

2 x = 9 0^{\circ} - (x - 1 5^{\circ}) + 36 0^{\circ} n : x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

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

Expand

9 0^{\circ} - (x - 1 5^{\circ}) + 36 0^{\circ} n : - x + 36 0^{\circ} n + 10 5^{\circ}

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

= - x + 1 5^{\circ}

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

Simplify

9 0^{\circ} - x + 1 5^{\circ} + 36 0^{\circ} n : - x + 36 0^{\circ} n + 10 5^{\circ}

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

Group like terms

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

Least Common Multiplier of

2, 12 : 12

2, 12

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

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

2

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

9 0^{\circ} :

multiply the denominator and numerator by

6

9 0^{\circ} = \frac{18 0 ^{\circ} 6}{2 \cdot 6} = 9 0^{\circ}

= 9 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} 6 + 18 0 ^{\circ}}{12}

Add similar elements:

108 0^{\circ} + 18 0^{\circ} = 126 0^{\circ}

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

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

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

Move

x

to the left side

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

Add

x

to both sides

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

Simplify

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

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

Divide both sides by

3

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

Divide both sides by

3

\frac{3 x}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{10 5 ^{\circ}}{3}

Simplify

\frac{3 x}{3} = \frac{36 0 ^{\circ} n}{3} + \frac{10 5 ^{\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{10 5 ^{\circ}}{3} : \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

\frac{36 0 ^{\circ} n}{3} + \frac{10 5 ^{\circ}}{3}

Apply rule

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

= \frac{36 0 ^{\circ} n + 10 5 ^{\circ}}{3}

Join

36 0^{\circ} n + 10 5^{\circ} : \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{12}

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

Convert element to fraction:

36 0^{\circ} n = \frac{36 0 ^{\circ} n 12}{12}

= \frac{36 0 ^{\circ} n \cdot 12}{12} + 10 5^{\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 12 + 126 0 ^{\circ}}{12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{12}

= \frac{\frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{12}}{3}

Apply the fraction rule:

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

= \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{12 \cdot 3}

Multiply the numbers:

12 \cdot 3 = 36

= \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

2 x = 18 0^{\circ} - (9 0^{\circ} - (x - 1 5^{\circ})) + 36 0^{\circ} n : x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

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

Expand

18 0^{\circ} - (9 0^{\circ} - (x - 1 5^{\circ})) + 36 0^{\circ} n : 18 0^{\circ} + x - 10 5^{\circ} + 36 0^{\circ} n

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

Expand

9 0^{\circ} - (x - 1 5^{\circ}) : - x + 10 5^{\circ}

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

= - x + 1 5^{\circ}

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

Simplify

9 0^{\circ} - x + 1 5^{\circ} : - x + 10 5^{\circ}

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

Group like terms

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

Least Common Multiplier of

2, 12 : 12

2, 12

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

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

2

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

9 0^{\circ} :

multiply the denominator and numerator by

6

9 0^{\circ} = \frac{18 0 ^{\circ} 6}{2 \cdot 6} = 9 0^{\circ}

= 9 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} 6 + 18 0 ^{\circ}}{12}

Add similar elements:

108 0^{\circ} + 18 0^{\circ} = 126 0^{\circ}

= - x + 10 5^{\circ}

= - x + 10 5^{\circ}

= 18 0^{\circ} - (- x + 10 5^{\circ}) + 36 0^{\circ} n

- (- x + 10 5^{\circ}) : x - 10 5^{\circ}

- (- x + 10 5^{\circ})

Distribute parentheses

= - (- x) - (10 5^{\circ})

Apply minus-plus rules

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

= x - 10 5^{\circ}

= 18 0^{\circ} + x - 10 5^{\circ} + 36 0^{\circ} n

2 x = 18 0^{\circ} + x - 10 5^{\circ} + 36 0^{\circ} n

Move

x

to the left side

2 x = 18 0^{\circ} + x - 10 5^{\circ} + 36 0^{\circ} n

Subtract

x

from both sides

2 x - x = 18 0^{\circ} + x - 10 5^{\circ} + 36 0^{\circ} n - x

Simplify

x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}, x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}, x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

Popular Examples

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tan (x^{2}) = 1

Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)=cos(x-15) ?
The general solution for sin(2x)=cos(x-15) is x=(4320n+1260)/(36),x=180-105+360n