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2/(sqrt(3))cos(3θ)=1,0<= θ<= 2pi

Solution

\frac{2}{3} cos (3 θ) = 1, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{18}, θ = \frac{11 π}{18}, θ = \frac{13 π}{18}, θ = \frac{23 π}{18}, θ = \frac{25 π}{18}, θ = \frac{35 π}{18}

Solution steps

\frac{2}{3} cos (3 θ) = 1, 0 \leq θ \leq 2 π

Simplify

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

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

\frac{2 3}{3} cos (3 θ) = 1

Multiply both sides by

3

\frac{2 3}{3} cos (3 θ) = 1

Multiply both sides by

3

3 \cdot \frac{2 3}{3} cos (3 θ) = 1 \cdot 3

Simplify

23 cos (3 θ) = 3

23 cos (3 θ) = 3

Divide both sides by

23

23 cos (3 θ) = 3

Divide both sides by

23

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

Simplify

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

Simplify

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

\frac{2 3 cos ( 3 θ )}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \frac{3 cos ( 3 θ )}{3}

Cancel the common factor:

3

= cos (3 θ)

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 (3 θ) = \frac{3}{2}

cos (3 θ) = \frac{3}{2}

cos (3 θ) = \frac{3}{2}

General solutions for

cos (3 θ) = \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}

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

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

Solve

3 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{18} + \frac{2 πn}{3}

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{6}}{3} + \frac{2 πn}{3} : \frac{π}{18} + \frac{2 πn}{3}

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

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{18} + \frac{2 πn}{3}

3 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3} : \frac{11 π}{18} + \frac{2 πn}{3}

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

\frac{\frac{11 π}{6}}{3} = \frac{11 π}{18}

\frac{\frac{11 π}{6}}{3}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{11 π}{18}

= \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{11 π}{18} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{18}, θ = \frac{11 π}{18}, θ = \frac{13 π}{18}, θ = \frac{23 π}{18}, θ = \frac{25 π}{18}, θ = \frac{35 π}{18}

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