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

The general solution for cos(2x)-cos(x)=1 is x=2.46670…+2pin,x=-2.46670…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 2.46670 \dots + 2 πn, x = - 2.46670 \dots + 2 πn

Degrees

x = 141.33171 \dots^{\circ} + 36 0^{\circ} n, x = - 141.33171 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

- (- a) = a

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

(- 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 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

= \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} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 2.46670 \dots + 2 πn, x = - 2.46670 \dots + 2 πn

Popular Examples

1-cot(x)=csc(x)

1 - cot (x) = csc (x)

2csc^2(x)=cot(x)

2 csc^{2} (x) = cot (x)

6sin(x)cos(x)+3sin(x)-2cos(x)-1=0

6 sin (x) cos (x) + 3 sin (x) - 2 cos (x) - 1 = 0

acos(x)+bsin(x)=0

a cos (x) + b sin (x) = 0

cot(x)= 7/4 ,sin(x)<0

cot (x) = \frac{7}{4}, sin (x) < 0

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-cos(x)=1 ?
The general solution for cos(2x)-cos(x)=1 is x=2.46670…+2pin,x=-2.46670…+2pin