What is the general solution for cos(5x)+cos(x)=2cos(2x) ?

The general solution for cos(5x)+cos(x)=2cos(2x) is x= pi/4+pin,x=(3pi)/4+pin,x=(2pin)/3

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Popular Trigonometry >

cos(5x)+cos(x)=2cos(2x)

Solution

cos (5 x) + cos (x) = 2 cos (2 x)

Solution

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

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 12 0^{\circ} n

Solution steps

cos (5 x) + cos (x) = 2 cos (2 x)

Subtract

2 cos (2 x)

from both sides

cos (5 x) + cos (x) - 2 cos (2 x) = 0

Rewrite using trig identities

cos (5 x) + cos (x) - 2 cos (2 x)

Use the Sum to Product identity:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= - 2 cos (2 x) + 2 cos (\frac{5 x + x}{2}) cos (\frac{5 x - x}{2})

2 cos (\frac{5 x + x}{2}) cos (\frac{5 x - x}{2}) = 2 cos (3 x) cos (2 x)

2 cos (\frac{5 x + x}{2}) cos (\frac{5 x - x}{2})

\frac{5 x + x}{2} = 3 x

\frac{5 x + x}{2}

Add similar elements:

5 x + x = 6 x

= \frac{6 x}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 x

= 2 cos (3 x) cos (\frac{5 x - x}{2})

\frac{5 x - x}{2} = 2 x

\frac{5 x - x}{2}

Add similar elements:

5 x - x = 4 x

= \frac{4 x}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 x

= 2 cos (3 x) cos (2 x)

= - 2 cos (2 x) + 2 cos (3 x) cos (2 x)

- 2 cos (2 x) + 2 cos (2 x) cos (3 x) = 0

Factor

- 2 cos (2 x) + 2 cos (2 x) cos (3 x) : 2 cos (2 x) (cos (3 x) - 1)

- 2 cos (2 x) + 2 cos (2 x) cos (3 x)

Factor out common term

2 cos (2 x)

= 2 cos (2 x) (- 1 + cos (3 x))

2 cos (2 x) (cos (3 x) - 1) = 0

Solving each part separately

cos (2 x) = 0 or cos (3 x) - 1 = 0

cos (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (2 x) = 0

General solutions for

cos (2 x) = 0

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}

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

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (3 x) - 1 = 0 : x = \frac{2 πn}{3}

cos (3 x) - 1 = 0

Move

1

to the right side

cos (3 x) - 1 = 0

Add

1

to both sides

cos (3 x) - 1 + 1 = 0 + 1

Simplify

cos (3 x) = 1

cos (3 x) = 1

General solutions for

cos (3 x) = 1

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 x = 0 + 2 πn

3 x = 0 + 2 πn

Solve

3 x = 0 + 2 πn : x = \frac{2 πn}{3}

3 x = 0 + 2 πn

0 + 2 πn = 2 πn

3 x = 2 πn

Divide both sides by

3

3 x = 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{2 πn}{3}

Simplify

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(5x)+cos(x)=2cos(2x) ?
The general solution for cos(5x)+cos(x)=2cos(2x) is x= pi/4+pin,x=(3pi)/4+pin,x=(2pin)/3