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

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

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

Solution

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

Solution

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

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

5 cos (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= - 5 (2 cos^{2} (x) - 1) + 5 cos (x)

- (- 1 + 2 cos^{2} (x)) \cdot 5 + 5 cos (x) = 0

Solve by substitution

- (- 1 + 2 cos^{2} (x)) \cdot 5 + 5 cos (x) = 0

Let:

cos (x) = u

- (- 1 + 2 u^{2}) \cdot 5 + 5 u = 0

- (- 1 + 2 u^{2}) \cdot 5 + 5 u = 0 : u = - \frac{1}{2}, u = 1

- (- 1 + 2 u^{2}) \cdot 5 + 5 u = 0

Expand

- (- 1 + 2 u^{2}) \cdot 5 + 5 u : 5 - 10 u^{2} + 5 u

- (- 1 + 2 u^{2}) \cdot 5 + 5 u

= - 5 (- 1 + 2 u^{2}) + 5 u

Expand

- 5 (- 1 + 2 u^{2}) : 5 - 10 u^{2}

- 5 (- 1 + 2 u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = - 5, b = - 1, c = 2 u^{2}

= - 5 (- 1) + (- 5) \cdot 2 u^{2}

Apply minus-plus rules

- (- a) = a, + (- a) = - a

= 5 \cdot 1 - 5 \cdot 2 u^{2}

Simplify

5 \cdot 1 - 5 \cdot 2 u^{2} : 5 - 10 u^{2}

5 \cdot 1 - 5 \cdot 2 u^{2}

Multiply the numbers:

5 \cdot 1 = 5

= 5 - 5 \cdot 2 u^{2}

Multiply the numbers:

5 \cdot 2 = 10

= 5 - 10 u^{2}

= 5 - 10 u^{2}

= 5 - 10 u^{2} + 5 u

5 - 10 u^{2} + 5 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 10 u^{2} + 5 u + 5 = 0

Solve with the quadratic formula

- 10 u^{2} + 5 u + 5 = 0

Quadratic Equation Formula:

For

a = - 10, b = 5, c = 5

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

5^{2} - 4 (- 10) \cdot 5 = 15

5^{2} - 4 (- 10) \cdot 5

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 5^{2} + 200

5^{2} = 25

= 25 + 200

Add the numbers:

25 + 200 = 225

= 225

Factor the number:

225 = 1 5^{2}

= 1 5^{2}

Apply radical rule:

1 5^{2} = 15

= 15

u_{1, 2} = \frac{- 5 \pm 15}{2 ( - 10 )}

Separate the solutions

u_{1} = \frac{- 5 + 15}{2 ( - 10 )}, u_{2} = \frac{- 5 - 15}{2 ( - 10 )}

u = \frac{- 5 + 15}{2 ( - 10 )} : - \frac{1}{2}

\frac{- 5 + 15}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 15}{- 2 \cdot 10}

Add/Subtract the numbers:

- 5 + 15 = 10

= \frac{10}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{10}{- 20}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{10}{20}

Cancel the common factor:

10

= - \frac{1}{2}

u = \frac{- 5 - 15}{2 ( - 10 )} : 1

\frac{- 5 - 15}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 15}{- 2 \cdot 10}

Subtract the numbers:

- 5 - 15 = - 20

= \frac{- 20}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 20}{- 20}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = 1

cos (x) = - \frac{1}{2}, cos (x) = 1

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

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

General solutions for

cos (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}

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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