What is the general solution for sin(x)-cos(x)= 1/(sqrt(2)) ?

The general solution for sin(x)-cos(x)= 1/(sqrt(2)) is x=2pin+(5pi)/(12),x=2pin+(13pi)/(12)

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sin(x)-cos(x)= 1/(sqrt(2))

Solution

sin (x) - cos (x) = \frac{1}{2}

Solution

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

Degrees

x = 7 5^{\circ} + 36 0^{\circ} n, x = 19 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) - cos (x) = \frac{1}{2}

Rewrite using trig identities

sin (x) - cos (x)

sin (x) - cos (x) = 2 sin (x - \frac{π}{4})

sin (x) - cos (x)

Rewrite as

= 2 (\frac{1}{2} sin (x) - \frac{1}{2} cos (x))

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) - sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (x - \frac{π}{4})

= 2 sin (x - \frac{π}{4})

2 sin (x - \frac{π}{4}) = \frac{1}{2}

Divide both sides by

2

2 sin (x - \frac{π}{4}) = \frac{1}{2}

Divide both sides by

2

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{\frac{1}{2}}{2}

Simplify

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{\frac{1}{2}}{2}

Simplify

\frac{2 sin ( x - \frac{π}{4} )}{2} : sin (x - \frac{π}{4})

\frac{2 sin ( x - \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x - \frac{π}{4})

Simplify

\frac{\frac{1}{2}}{2} : \frac{1}{2}

\frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

= \frac{1}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

General solutions for

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x - \frac{π}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

x - \frac{π}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

Solve

x - \frac{π}{4} = \frac{π}{6} + 2 πn : x = 2 πn + \frac{5 π}{12}

x - \frac{π}{4} = \frac{π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

x - \frac{π}{4} = \frac{π}{6} + 2 πn

Add

\frac{π}{4}

to both sides

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

Simplify

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

Simplify

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= x

Simplify

\frac{π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{5 π}{12}

\frac{π}{6} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{6} + \frac{π}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

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

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

6

4

= 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

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 2}{12} + \frac{π 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 + π 3}{12}

Add similar elements:

2 π + 3 π = 5 π

= 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

Solve

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn : x = 2 πn + \frac{13 π}{12}

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

Add

\frac{π}{4}

to both sides

x - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{6} + 2 πn + \frac{π}{4}

Simplify

x - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{6} + 2 πn + \frac{π}{4}

Simplify

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= x

Simplify

\frac{5 π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{13 π}{12}

\frac{5 π}{6} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{5 π}{6}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

4

6

= 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

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

For

\frac{5 π}{6} :

multiply the denominator and numerator by

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= \frac{π 3}{12} + \frac{10 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 3 + 10 π}{12}

Add similar elements:

3 π + 10 π = 13 π

= 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

Popular Examples

cos(x)= 6/15

cos (x) = \frac{6}{15}

arccos(x)=90

arccos (x) = 90

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

2 sin (2 x) = sin^{2} (x) + cos^{2} (x)

sin(θ)= 7/12

sin (θ) = \frac{7}{12}

3cos(2x)+2sin^2(x)=2

3 cos (2 x) + 2 sin^{2} (x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-cos(x)= 1/(sqrt(2)) ?
The general solution for sin(x)-cos(x)= 1/(sqrt(2)) is x=2pin+(5pi)/(12),x=2pin+(13pi)/(12)