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

The general solution for cos(2x)+cos(x)-1=0 is x=0.67488…+2pin,x=2pi-0.67488…+2pin

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

Solution

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

Solution

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

Degrees

x = 38.66828 \dots^{\circ} + 36 0^{\circ} n, x = 321.33171 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- 2 + u + 2 u^{2} = 0 : u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 2 (- 2) = 17

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 17}{2 \cdot 2}, u_{2} = \frac{- 1 - 17}{2 \cdot 2}

u = \frac{- 1 + 17}{2 \cdot 2} : \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 \cdot 2} : \frac{- 1 - 17}{4}

\frac{- 1 - 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 17}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

Substitute back

u = cos (x)

cos (x) = \frac{- 1 + 17}{4}, cos (x) = \frac{- 1 - 17}{4}

cos (x) = \frac{- 1 + 17}{4}, cos (x) = \frac{- 1 - 17}{4}

cos (x) = \frac{- 1 + 17}{4} : x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

cos (x) = \frac{- 1 + 17}{4}

Apply trig inverse properties

cos (x) = \frac{- 1 + 17}{4}

General solutions for

cos (x) = \frac{- 1 + 17}{4}

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

x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

cos (x) = \frac{- 1 - 17}{4} :

No Solution

cos (x) = \frac{- 1 - 17}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

Show solutions in decimal form

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)+cos(x)-1=0 ?
The general solution for cos(2x)+cos(x)-1=0 is x=0.67488…+2pin,x=2pi-0.67488…+2pin