What is the general solution for 4cos(2x)=1-3cos(x) ?

The general solution for 4cos(2x)=1-3cos(x) is x=0.89566…+2pin,x=2pi-0.89566…+2pin,x=pi+2pin

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4cos(2x)=1-3cos(x)

Solution

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

Solution

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

Degrees

x = 51.31781 \dots^{\circ} + 36 0^{\circ} n, x = 308.68218 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - 3 cos (x)

from both sides

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

Rewrite using trig identities

- 1 + 3 cos (x) + 4 cos (2 x)

Use the Double Angle identity:

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

= - 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1)

Simplify

- 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) + 3 cos (x) - 5

- 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1)

Expand

4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 4

4 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

= 4 \cdot 2 cos^{2} (x) - 4 \cdot 1

Simplify

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

4 \cdot 2 cos^{2} (x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= - 1 + 3 cos (x) + 8 cos^{2} (x) - 4

Simplify

- 1 + 3 cos (x) + 8 cos^{2} (x) - 4 : 8 cos^{2} (x) + 3 cos (x) - 5

- 1 + 3 cos (x) + 8 cos^{2} (x) - 4

Group like terms

= 3 cos (x) + 8 cos^{2} (x) - 1 - 4

Subtract the numbers:

- 1 - 4 = - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

- 5 + 3 cos (x) + 8 cos^{2} (x) = 0

Solve by substitution

- 5 + 3 cos (x) + 8 cos^{2} (x) = 0

Let:

cos (x) = u

- 5 + 3 u + 8 u^{2} = 0

- 5 + 3 u + 8 u^{2} = 0 : u = \frac{5}{8}, u = - 1

- 5 + 3 u + 8 u^{2} = 0

Write in the standard form

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

8 u^{2} + 3 u - 5 = 0

Solve with the quadratic formula

8 u^{2} + 3 u - 5 = 0

Quadratic Equation Formula:

For

a = 8, b = 3, c = - 5

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

3^{2} - 4 \cdot 8 (- 5) = 13

3^{2} - 4 \cdot 8 (- 5)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 8 \cdot 5

Multiply the numbers:

4 \cdot 8 \cdot 5 = 160

= 3^{2} + 160

3^{2} = 9

= 9 + 160

Add the numbers:

9 + 160 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 3 + 13}{2 \cdot 8}, u_{2} = \frac{- 3 - 13}{2 \cdot 8}

u = \frac{- 3 + 13}{2 \cdot 8} : \frac{5}{8}

\frac{- 3 + 13}{2 \cdot 8}

Add/Subtract the numbers:

- 3 + 13 = 10

= \frac{10}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{10}{16}

Cancel the common factor:

2

= \frac{5}{8}

u = \frac{- 3 - 13}{2 \cdot 8} : - 1

\frac{- 3 - 13}{2 \cdot 8}

Subtract the numbers:

- 3 - 13 = - 16

= \frac{- 16}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 16}{16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{5}{8}, u = - 1

Substitute back

u = cos (x)

cos (x) = \frac{5}{8}, cos (x) = - 1

cos (x) = \frac{5}{8}, cos (x) = - 1

cos (x) = \frac{5}{8} : x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

cos (x) = \frac{5}{8}

Apply trig inverse properties

cos (x) = \frac{5}{8}

General solutions for

cos (x) = \frac{5}{8}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn, x = π + 2 πn

Show solutions in decimal form

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2x)=1-3cos(x) ?
The general solution for 4cos(2x)=1-3cos(x) is x=0.89566…+2pin,x=2pi-0.89566…+2pin,x=pi+2pin