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Popular >

-2cos(3θ)=1,0<= θ<= 2pi

Solution

- 2 cos (3 θ) = 1, 0 \leq θ \leq 2 π

Solution

θ = \frac{2 π}{9}, θ = \frac{4 π}{9}, θ = \frac{8 π}{9}, θ = \frac{10 π}{9}, θ = \frac{14 π}{9}, θ = \frac{16 π}{9}

+1

Degrees

θ = 4 0^{\circ}, θ = 8 0^{\circ}, θ = 16 0^{\circ}, θ = 20 0^{\circ}, θ = 28 0^{\circ}, θ = 32 0^{\circ}

Solution steps

- 2 cos (3 θ) = 1, 0 \leq θ \leq 2 π

Divide both sides by

- 2

- 2 cos (3 θ) = 1

Divide both sides by

- 2

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

Simplify

cos (3 θ) = - \frac{1}{2}

cos (3 θ) = - \frac{1}{2}

General solutions for

cos (3 θ) = - \frac{1}{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{2 π}{3} + 2 πn, 3 θ = \frac{4 π}{3} + 2 πn

3 θ = \frac{2 π}{3} + 2 πn, 3 θ = \frac{4 π}{3} + 2 πn

Solve

3 θ = \frac{2 π}{3} + 2 πn : θ = \frac{2 π}{9} + \frac{2 πn}{3}

3 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

3

3 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

\frac{3 θ}{3} = \frac{\frac{2 π}{3}}{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{2 π}{3}}{3} + \frac{2 πn}{3} : \frac{2 π}{9} + \frac{2 πn}{3}

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= \frac{2 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}

Solve

3 θ = \frac{4 π}{3} + 2 πn : θ = \frac{4 π}{9} + \frac{2 πn}{3}

3 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

3

3 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{4 π}{3}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{4 π}{3}}{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{4 π}{3}}{3} + \frac{2 πn}{3} : \frac{4 π}{9} + \frac{2 πn}{3}

\frac{\frac{4 π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{4 π}{3}}{3} = \frac{4 π}{9}

\frac{\frac{4 π}{3}}{3}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{4 π}{9}

= \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}, θ = \frac{4 π}{9} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{2 π}{9}, θ = \frac{4 π}{9}, θ = \frac{8 π}{9}, θ = \frac{10 π}{9}, θ = \frac{14 π}{9}, θ = \frac{16 π}{9}

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