What is the general solution for 2cos(2x)-cos(x)-1=0 ?

The general solution for 2cos(2x)-cos(x)-1=0 is x=2pin,x=2.41885…+2pin,x=-2.41885…+2pin

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

Solution

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

Solution

x = 2 πn, x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 2 = - 3

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

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

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

- 3 - cos (x) + 4 cos^{2} (x) = 0

Solve by substitution

- 3 - cos (x) + 4 cos^{2} (x) = 0

Let:

cos (x) = u

- 3 - u + 4 u^{2} = 0

- 3 - u + 4 u^{2} = 0 : u = 1, u = - \frac{3}{4}

- 3 - u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} - u - 3 = 0

Solve with the quadratic formula

4 u^{2} - u - 3 = 0

Quadratic Equation Formula:

For

a = 4, b = - 1, c = - 3

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

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

(- 1)^{2} - 4 \cdot 4 (- 3) = 7

(- 1)^{2} - 4 \cdot 4 (- 3)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \cdot 3

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 4}

u = \frac{- ( - 1 ) + 7}{2 \cdot 4} : 1

\frac{- ( - 1 ) + 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 4}

Add the numbers:

1 + 7 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 7}{2 \cdot 4} : - \frac{3}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 4}

Subtract the numbers:

1 - 7 = - 6

= \frac{- 6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{8}

Apply the fraction rule:

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

= - \frac{6}{8}

Cancel the common factor:

2

= - \frac{3}{4}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{3}{4}

cos (x) = 1, cos (x) = - \frac{3}{4}

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

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

cos (x) = - \frac{3}{4}

Apply trig inverse properties

cos (x) = - \frac{3}{4}

General solutions for

cos (x) = - \frac{3}{4}

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

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)-cos(x)-1=0 ?
The general solution for 2cos(2x)-cos(x)-1=0 is x=2pin,x=2.41885…+2pin,x=-2.41885…+2pin