What is the general solution for 2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi ?

Q: What is the general solution for 2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi ?

The general solution for 2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi is θ= pi/2 ,θ= pi/6

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Popular Trigonometry >

2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi

Solution

23 cos (θ - \frac{π}{3}) = 3, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{2}, θ = \frac{π}{6}

Degrees

θ = 9 0^{\circ}, θ = 3 0^{\circ}

Solution steps

23 cos (θ - \frac{π}{3}) = 3, 0 \leq θ \leq 2 π

Divide both sides by

23

23 cos (θ - \frac{π}{3}) = 3

Divide both sides by

23

\frac{2 3 cos ( θ - \frac{π}{3} )}{2 3} = \frac{3}{2 3}

Simplify

\frac{2 3 cos ( θ - \frac{π}{3} )}{2 3} = \frac{3}{2 3}

Simplify

\frac{2 3 cos ( θ - \frac{π}{3} )}{2 3} : cos (θ - \frac{π}{3})

\frac{2 3 cos ( θ - \frac{π}{3} )}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \frac{3 cos ( θ - \frac{π}{3} )}{3}

Cancel the common factor:

3

= cos (θ - \frac{π}{3})

Simplify

\frac{3}{2 3} : \frac{3}{2}

\frac{3}{2 3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{2 \cdot 3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= \frac{3 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{3}{2}

cos (θ - \frac{π}{3}) = \frac{3}{2}

cos (θ - \frac{π}{3}) = \frac{3}{2}

cos (θ - \frac{π}{3}) = \frac{3}{2}

General solutions for

cos (θ - \frac{π}{3}) = \frac{3}{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}

θ - \frac{π}{3} = \frac{π}{6} + 2 πn, θ - \frac{π}{3} = \frac{11 π}{6} + 2 πn

θ - \frac{π}{3} = \frac{π}{6} + 2 πn, θ - \frac{π}{3} = \frac{11 π}{6} + 2 πn

Solve

θ - \frac{π}{3} = \frac{π}{6} + 2 πn : θ = 2 πn + \frac{π}{2}

θ - \frac{π}{3} = \frac{π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

θ - \frac{π}{3} = \frac{π}{6} + 2 πn

Add

\frac{π}{3}

to both sides

θ - \frac{π}{3} + \frac{π}{3} = \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

θ - \frac{π}{3} + \frac{π}{3} = \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

θ - \frac{π}{3} + \frac{π}{3} : θ

θ - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} = 0

= θ

Simplify

\frac{π}{6} + 2 πn + \frac{π}{3} : 2 πn + \frac{π}{2}

\frac{π}{6} + 2 πn + \frac{π}{3}

Group like terms

= 2 πn + \frac{π}{6} + \frac{π}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

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

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

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= \frac{π}{6} + \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π + π 2}{6}

Add similar elements:

π + 2 π = 3 π

= \frac{3 π}{6}

Cancel the common factor:

3

= 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

Solve

θ - \frac{π}{3} = \frac{11 π}{6} + 2 πn : θ = 2 πn + \frac{13 π}{6}

θ - \frac{π}{3} = \frac{11 π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

θ - \frac{π}{3} = \frac{11 π}{6} + 2 πn

Add

\frac{π}{3}

to both sides

θ - \frac{π}{3} + \frac{π}{3} = \frac{11 π}{6} + 2 πn + \frac{π}{3}

Simplify

θ - \frac{π}{3} + \frac{π}{3} = \frac{11 π}{6} + 2 πn + \frac{π}{3}

Simplify

θ - \frac{π}{3} + \frac{π}{3} : θ

θ - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} = 0

= θ

Simplify

\frac{11 π}{6} + 2 πn + \frac{π}{3} : 2 πn + \frac{13 π}{6}

\frac{11 π}{6} + 2 πn + \frac{π}{3}

Group like terms

= 2 πn + \frac{π}{3} + \frac{11 π}{6}

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= \frac{π 2}{6} + \frac{11 π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 + 11 π}{6}

Add similar elements:

2 π + 11 π = 13 π

= 2 πn + \frac{13 π}{6}

θ = 2 πn + \frac{13 π}{6}

θ = 2 πn + \frac{13 π}{6}

θ = 2 πn + \frac{13 π}{6}

θ = 2 πn + \frac{π}{2}, θ = 2 πn + \frac{13 π}{6}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{π}{6}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi ?
The general solution for 2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi is θ= pi/2 ,θ= pi/6