What is the general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi ?

Q: What is the general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi ?

The general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi is x=(25pi)/(28),x=(39pi)/(28)

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cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi

Solution

cos (x - \frac{π}{7}) = \frac{- 2}{2}, 0 \leq x \leq 2 π

Solution

x = \frac{25 π}{28}, x = \frac{39 π}{28}

Degrees

x = 160.71428 \dots^{\circ}, x = 250.71428 \dots^{\circ}

Solution steps

cos (x - \frac{π}{7}) = \frac{- 2}{2}, 0 \leq x \leq 2 π

Apply the fraction rule:

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

cos (x - \frac{π}{7}) = - \frac{2}{2}

General solutions for

cos (x - \frac{π}{7}) = - \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn, x - \frac{π}{7} = \frac{5 π}{4} + 2 πn

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn, x - \frac{π}{7} = \frac{5 π}{4} + 2 πn

Solve

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn : x = 2 πn + \frac{25 π}{28}

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn

Move

\frac{π}{7}

to the right side

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn

Add

\frac{π}{7}

to both sides

x - \frac{π}{7} + \frac{π}{7} = \frac{3 π}{4} + 2 πn + \frac{π}{7}

Simplify

x - \frac{π}{7} + \frac{π}{7} = \frac{3 π}{4} + 2 πn + \frac{π}{7}

Simplify

x - \frac{π}{7} + \frac{π}{7} : x

x - \frac{π}{7} + \frac{π}{7}

Add similar elements:

- \frac{π}{7} + \frac{π}{7} = 0

= x

Simplify

\frac{3 π}{4} + 2 πn + \frac{π}{7} : 2 πn + \frac{25 π}{28}

\frac{3 π}{4} + 2 πn + \frac{π}{7}

Group like terms

= 2 πn + \frac{π}{7} + \frac{3 π}{4}

Least Common Multiplier of

7, 4 : 28

7, 4

Least Common Multiplier (LCM)

Prime factorization of

7 : 7

7

7

is a prime number, therefore no factorization is possible

= 7

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

7

4

= 7 \cdot 2 \cdot 2

Multiply the numbers:

7 \cdot 2 \cdot 2 = 28

= 28

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

28

For

\frac{π}{7} :

multiply the denominator and numerator by

4

\frac{π}{7} = \frac{π 4}{7 \cdot 4} = \frac{π 4}{28}

For

\frac{3 π}{4} :

multiply the denominator and numerator by

7

\frac{3 π}{4} = \frac{3 π 7}{4 \cdot 7} = \frac{21 π}{28}

= \frac{π 4}{28} + \frac{21 π}{28}

Since the denominators are equal, combine the fractions:

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

= \frac{π 4 + 21 π}{28}

Add similar elements:

4 π + 21 π = 25 π

= 2 πn + \frac{25 π}{28}

x = 2 πn + \frac{25 π}{28}

x = 2 πn + \frac{25 π}{28}

x = 2 πn + \frac{25 π}{28}

Solve

x - \frac{π}{7} = \frac{5 π}{4} + 2 πn : x = 2 πn + \frac{39 π}{28}

x - \frac{π}{7} = \frac{5 π}{4} + 2 πn

Move

\frac{π}{7}

to the right side

x - \frac{π}{7} = \frac{5 π}{4} + 2 πn

Add

\frac{π}{7}

to both sides

x - \frac{π}{7} + \frac{π}{7} = \frac{5 π}{4} + 2 πn + \frac{π}{7}

Simplify

x - \frac{π}{7} + \frac{π}{7} = \frac{5 π}{4} + 2 πn + \frac{π}{7}

Simplify

x - \frac{π}{7} + \frac{π}{7} : x

x - \frac{π}{7} + \frac{π}{7}

Add similar elements:

- \frac{π}{7} + \frac{π}{7} = 0

= x

Simplify

\frac{5 π}{4} + 2 πn + \frac{π}{7} : 2 πn + \frac{39 π}{28}

\frac{5 π}{4} + 2 πn + \frac{π}{7}

Group like terms

= 2 πn + \frac{π}{7} + \frac{5 π}{4}

Least Common Multiplier of

7, 4 : 28

7, 4

Least Common Multiplier (LCM)

Prime factorization of

7 : 7

7

7

is a prime number, therefore no factorization is possible

= 7

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

7

4

= 7 \cdot 2 \cdot 2

Multiply the numbers:

7 \cdot 2 \cdot 2 = 28

= 28

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

28

For

\frac{π}{7} :

multiply the denominator and numerator by

4

\frac{π}{7} = \frac{π 4}{7 \cdot 4} = \frac{π 4}{28}

For

\frac{5 π}{4} :

multiply the denominator and numerator by

7

\frac{5 π}{4} = \frac{5 π 7}{4 \cdot 7} = \frac{35 π}{28}

= \frac{π 4}{28} + \frac{35 π}{28}

Since the denominators are equal, combine the fractions:

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

= \frac{π 4 + 35 π}{28}

Add similar elements:

4 π + 35 π = 39 π

= 2 πn + \frac{39 π}{28}

x = 2 πn + \frac{39 π}{28}

x = 2 πn + \frac{39 π}{28}

x = 2 πn + \frac{39 π}{28}

x = 2 πn + \frac{25 π}{28}, x = 2 πn + \frac{39 π}{28}

Solutions for the range

0 \leq x \leq 2 π

x = \frac{25 π}{28}, x = \frac{39 π}{28}

Popular Examples

sin(2θ)=sin(θ)

sin (2 θ) = sin (θ)

25cos^2(θ)-1=0

25 cos^{2} (θ) - 1 = 0

cos(θ)= 4/7

cos (θ) = \frac{4}{7}

cos(θ)-1=-1

cos (θ) - 1 = - 1

2=sec(θ)

2 = sec (θ)

Frequently Asked Questions (FAQ)

What is the general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi ?
The general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi is x=(25pi)/(28),x=(39pi)/(28)