What is the general solution for 2sin(pi/4-x)+sqrt(3)=0 ?

The general solution for 2sin(pi/4-x)+sqrt(3)=0 is x=-2pin-(13pi)/(12),x=-2pin-(17pi)/(12)

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2sin(pi/4-x)+sqrt(3)=0

Solution

2 sin (\frac{π}{4} - x) + 3 = 0

Solution

x = - 2 πn - \frac{13 π}{12}, x = - 2 πn - \frac{17 π}{12}

Degrees

x = - 19 5^{\circ} - 36 0^{\circ} n, x = - 25 5^{\circ} - 36 0^{\circ} n

Solution steps

2 sin (\frac{π}{4} - x) + 3 = 0

Move

3

to the right side

2 sin (\frac{π}{4} - x) + 3 = 0

Subtract

3

from both sides

2 sin (\frac{π}{4} - x) + 3 - 3 = 0 - 3

Simplify

2 sin (\frac{π}{4} - x) = - 3

2 sin (\frac{π}{4} - x) = - 3

Divide both sides by

2

2 sin (\frac{π}{4} - x) = - 3

Divide both sides by

2

\frac{2 sin ( \frac{π}{4} - x )}{2} = \frac{- 3}{2}

Simplify

sin (\frac{π}{4} - x) = - \frac{3}{2}

sin (\frac{π}{4} - x) = - \frac{3}{2}

General solutions for

sin (\frac{π}{4} - x) = - \frac{3}{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}

\frac{π}{4} - x = \frac{4 π}{3} + 2 πn, \frac{π}{4} - x = \frac{5 π}{3} + 2 πn

\frac{π}{4} - x = \frac{4 π}{3} + 2 πn, \frac{π}{4} - x = \frac{5 π}{3} + 2 πn

Solve

\frac{π}{4} - x = \frac{4 π}{3} + 2 πn : x = - 2 πn - \frac{13 π}{12}

\frac{π}{4} - x = \frac{4 π}{3} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{π}{4} - x = \frac{4 π}{3} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} - x - \frac{π}{4} = \frac{4 π}{3} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} - x - \frac{π}{4} = \frac{4 π}{3} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} - x - \frac{π}{4} : - x

\frac{π}{4} - x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= - x

Simplify

\frac{4 π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{13 π}{12}

\frac{4 π}{3} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{4 π}{3}

Least Common Multiplier of

4, 3 : 12

4, 3

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

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

4

3

= 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{4 π}{3} :

multiply the denominator and numerator by

4

\frac{4 π}{3} = \frac{4 π 4}{3 \cdot 4} = \frac{16 π}{12}

= - \frac{π 3}{12} + \frac{16 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 + 16 π}{12}

Add similar elements:

- 3 π + 16 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}

Divide both sides by

- 1

- x = 2 πn + \frac{13 π}{12}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{13 π}{12}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{13 π}{12}}{- 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{2 πn}{- 1} + \frac{\frac{13 π}{12}}{- 1} : - 2 πn - \frac{13 π}{12}

\frac{2 πn}{- 1} + \frac{\frac{13 π}{12}}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn + \frac{\frac{13 π}{12}}{- 1}

\frac{\frac{13 π}{12}}{- 1} = - \frac{13 π}{12}

\frac{\frac{13 π}{12}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{13 π}{12}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{13 π}{12}}{1} = \frac{13 π}{12}

= - \frac{13 π}{12}

= - 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}

Solve

\frac{π}{4} - x = \frac{5 π}{3} + 2 πn : x = - 2 πn - \frac{17 π}{12}

\frac{π}{4} - x = \frac{5 π}{3} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{π}{4} - x = \frac{5 π}{3} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} - x - \frac{π}{4} = \frac{5 π}{3} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} - x - \frac{π}{4} = \frac{5 π}{3} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} - x - \frac{π}{4} : - x

\frac{π}{4} - x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= - x

Simplify

\frac{5 π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{17 π}{12}

\frac{5 π}{3} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{5 π}{3}

Least Common Multiplier of

4, 3 : 12

4, 3

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

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

4

3

= 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}{3} :

multiply the denominator and numerator by

4

\frac{5 π}{3} = \frac{5 π 4}{3 \cdot 4} = \frac{20 π}{12}

= - \frac{π 3}{12} + \frac{20 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 + 20 π}{12}

Add similar elements:

- 3 π + 20 π = 17 π

= 2 πn + \frac{17 π}{12}

- x = 2 πn + \frac{17 π}{12}

- x = 2 πn + \frac{17 π}{12}

- x = 2 πn + \frac{17 π}{12}

Divide both sides by

- 1

- x = 2 πn + \frac{17 π}{12}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{17 π}{12}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{17 π}{12}}{- 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{2 πn}{- 1} + \frac{\frac{17 π}{12}}{- 1} : - 2 πn - \frac{17 π}{12}

\frac{2 πn}{- 1} + \frac{\frac{17 π}{12}}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn + \frac{\frac{17 π}{12}}{- 1}

\frac{\frac{17 π}{12}}{- 1} = - \frac{17 π}{12}

\frac{\frac{17 π}{12}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{17 π}{12}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{17 π}{12}}{1} = \frac{17 π}{12}

= - \frac{17 π}{12}

= - 2 πn - \frac{17 π}{12}

x = - 2 πn - \frac{17 π}{12}

x = - 2 πn - \frac{17 π}{12}

x = - 2 πn - \frac{17 π}{12}

x = - 2 πn - \frac{13 π}{12}, x = - 2 πn - \frac{17 π}{12}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(pi/4-x)+sqrt(3)=0 ?
The general solution for 2sin(pi/4-x)+sqrt(3)=0 is x=-2pin-(13pi)/(12),x=-2pin-(17pi)/(12)