What is the general solution for 2cos(9x)=2-2cos(9x) ?

The general solution for 2cos(9x)=2-2cos(9x) is x= pi/(27)+(2pin)/9 ,x=(5pi}{27}+\frac{2pin)/9

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2cos(9x)=2-2cos(9x)

Solution

2 cos (9 x) = 2 - 2 cos (9 x)

Solution

x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

Degrees

x = 6.66666 \dots^{\circ} + 4 0^{\circ} n, x = 33.33333 \dots^{\circ} + 4 0^{\circ} n

Solution steps

2 cos (9 x) = 2 - 2 cos (9 x)

Solve by substitution

2 cos (9 x) = 2 - 2 cos (9 x)

Let:

cos (9 x) = u

2 u = 2 - 2 u

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

2 u = 2 - 2 u

Move

2 u

to the left side

2 u = 2 - 2 u

Add

2 u

to both sides

2 u + 2 u = 2 - 2 u + 2 u

Simplify

4 u = 2

4 u = 2

Divide both sides by

4

4 u = 2

Divide both sides by

4

\frac{4 u}{4} = \frac{2}{4}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (9 x)

cos (9 x) = \frac{1}{2}

cos (9 x) = \frac{1}{2}

cos (9 x) = \frac{1}{2} : x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

cos (9 x) = \frac{1}{2}

General solutions for

cos (9 x) = \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}

9 x = \frac{π}{3} + 2 πn, 9 x = \frac{5 π}{3} + 2 πn

9 x = \frac{π}{3} + 2 πn, 9 x = \frac{5 π}{3} + 2 πn

Solve

9 x = \frac{π}{3} + 2 πn : x = \frac{π}{27} + \frac{2 πn}{9}

9 x = \frac{π}{3} + 2 πn

Divide both sides by

9

9 x = \frac{π}{3} + 2 πn

Divide both sides by

9

\frac{9 x}{9} = \frac{\frac{π}{3}}{9} + \frac{2 πn}{9}

Simplify

\frac{9 x}{9} = \frac{\frac{π}{3}}{9} + \frac{2 πn}{9}

Simplify

\frac{9 x}{9} : x

\frac{9 x}{9}

Divide the numbers:

\frac{9}{9} = 1

= x

Simplify

\frac{\frac{π}{3}}{9} + \frac{2 πn}{9} : \frac{π}{27} + \frac{2 πn}{9}

\frac{\frac{π}{3}}{9} + \frac{2 πn}{9}

\frac{\frac{π}{3}}{9} = \frac{π}{27}

\frac{\frac{π}{3}}{9}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 9}

Multiply the numbers:

3 \cdot 9 = 27

= \frac{π}{27}

= \frac{π}{27} + \frac{2 πn}{9}

x = \frac{π}{27} + \frac{2 πn}{9}

x = \frac{π}{27} + \frac{2 πn}{9}

x = \frac{π}{27} + \frac{2 πn}{9}

Solve

9 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{27} + \frac{2 πn}{9}

9 x = \frac{5 π}{3} + 2 πn

Divide both sides by

9

9 x = \frac{5 π}{3} + 2 πn

Divide both sides by

9

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

Simplify

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

Simplify

\frac{9 x}{9} : x

\frac{9 x}{9}

Divide the numbers:

\frac{9}{9} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{9} + \frac{2 πn}{9} : \frac{5 π}{27} + \frac{2 πn}{9}

\frac{\frac{5 π}{3}}{9} + \frac{2 πn}{9}

\frac{\frac{5 π}{3}}{9} = \frac{5 π}{27}

\frac{\frac{5 π}{3}}{9}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 9}

Multiply the numbers:

3 \cdot 9 = 27

= \frac{5 π}{27}

= \frac{5 π}{27} + \frac{2 πn}{9}

x = \frac{5 π}{27} + \frac{2 πn}{9}

x = \frac{5 π}{27} + \frac{2 πn}{9}

x = \frac{5 π}{27} + \frac{2 πn}{9}

x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

Combine all the solutions

x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

Popular Examples

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sin (2 x) = 32, - π \leq x \leq π

5cos^2(x)-2=0

5 cos^{2} (x) - 2 = 0

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cos^{2} (x) + sec^{2} (x) = 11

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5 cos^{2} (x) - 3 = 0

csc(2x+3)= 5/4

csc (2 x + 3^{\circ}) = \frac{5}{4}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(9x)=2-2cos(9x) ?
The general solution for 2cos(9x)=2-2cos(9x) is x= pi/(27)+(2pin)/9 ,x=(5pi}{27}+\frac{2pin)/9