What is the general solution for 2cos((x+pi)/6)=1,0<x<2pi ?

Q: What is the general solution for 2cos((x+pi)/6)=1,0<x<2pi ?

The general solution for 2cos((x+pi)/6)=1,0<x<2pi is x=pi

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2cos((x+pi)/6)=1,0<x<2pi

Solution

2 cos (\frac{x + π}{6}) = 1, 0 < x < 2 π

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

2 cos (\frac{x + π}{6}) = 1, 0 < x < 2 π

Divide both sides by

2

2 cos (\frac{x + π}{6}) = 1

Divide both sides by

2

\frac{2 cos ( \frac{x + π}{6} )}{2} = \frac{1}{2}

Simplify

cos (\frac{x + π}{6}) = \frac{1}{2}

cos (\frac{x + π}{6}) = \frac{1}{2}

General solutions for

cos (\frac{x + π}{6}) = \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}

\frac{x + π}{6} = \frac{π}{3} + 2 πn, \frac{x + π}{6} = \frac{5 π}{3} + 2 πn

\frac{x + π}{6} = \frac{π}{3} + 2 πn, \frac{x + π}{6} = \frac{5 π}{3} + 2 πn

Solve

\frac{x + π}{6} = \frac{π}{3} + 2 πn : x = π + 12 πn

\frac{x + π}{6} = \frac{π}{3} + 2 πn

Multiply both sides by

6

\frac{x + π}{6} = \frac{π}{3} + 2 πn

Multiply both sides by

6

\frac{6 ( x + π )}{6} = 6 \cdot \frac{π}{3} + 6 \cdot 2 πn

Simplify

\frac{6 ( x + π )}{6} = 6 \cdot \frac{π}{3} + 6 \cdot 2 πn

Simplify

\frac{6 ( x + π )}{6} : x + π

\frac{6 ( x + π )}{6}

Divide the numbers:

\frac{6}{6} = 1

= x + π

Simplify

6 \cdot \frac{π}{3} + 6 \cdot 2 πn : 2 π + 12 πn

6 \cdot \frac{π}{3} + 6 \cdot 2 πn

6 \cdot \frac{π}{3} = 2 π

6 \cdot \frac{π}{3}

Multiply fractions:

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

= \frac{π 6}{3}

Divide the numbers:

\frac{6}{3} = 2

= 2 π

6 \cdot 2 πn = 12 πn

6 \cdot 2 πn

Multiply the numbers:

6 \cdot 2 = 12

= 12 πn

= 2 π + 12 πn

x + π = 2 π + 12 πn

x + π = 2 π + 12 πn

x + π = 2 π + 12 πn

Move

π

to the right side

x + π = 2 π + 12 πn

Subtract

π

from both sides

x + π - π = 2 π + 12 πn - π

Simplify

x = π + 12 πn

x = π + 12 πn

Solve

\frac{x + π}{6} = \frac{5 π}{3} + 2 πn : x = 9 π + 12 πn

\frac{x + π}{6} = \frac{5 π}{3} + 2 πn

Multiply both sides by

6

\frac{x + π}{6} = \frac{5 π}{3} + 2 πn

Multiply both sides by

6

\frac{6 ( x + π )}{6} = 6 \cdot \frac{5 π}{3} + 6 \cdot 2 πn

Simplify

\frac{6 ( x + π )}{6} = 6 \cdot \frac{5 π}{3} + 6 \cdot 2 πn

Simplify

\frac{6 ( x + π )}{6} : x + π

\frac{6 ( x + π )}{6}

Divide the numbers:

\frac{6}{6} = 1

= x + π

Simplify

6 \cdot \frac{5 π}{3} + 6 \cdot 2 πn : 10 π + 12 πn

6 \cdot \frac{5 π}{3} + 6 \cdot 2 πn

6 \cdot \frac{5 π}{3} = 10 π

6 \cdot \frac{5 π}{3}

Multiply fractions:

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

= \frac{5 π 6}{3}

Multiply the numbers:

5 \cdot 6 = 30

= \frac{30 π}{3}

Divide the numbers:

\frac{30}{3} = 10

= 10 π

6 \cdot 2 πn = 12 πn

6 \cdot 2 πn

Multiply the numbers:

6 \cdot 2 = 12

= 12 πn

= 10 π + 12 πn

x + π = 10 π + 12 πn

x + π = 10 π + 12 πn

x + π = 10 π + 12 πn

Move

π

to the right side

x + π = 10 π + 12 πn

Subtract

π

from both sides

x + π - π = 10 π + 12 πn - π

Simplify

x = 9 π + 12 πn

x = 9 π + 12 πn

x = π + 12 πn, x = 9 π + 12 πn

Solutions for the range

0 < x < 2 π

x = π

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos((x+pi)/6)=1,0<x<2pi ?
The general solution for 2cos((x+pi)/6)=1,0<x<2pi is x=pi