What is the general solution for cos(2x)-5cos(x)-1=0 ?

The general solution for cos(2x)-5cos(x)-1=0 is x=1.92920…+2pin,x=-1.92920…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2x)-5cos(x)-1=0

Solution

cos (2 x) - 5 cos (x) - 1 = 0

Solution

x = 1.92920 \dots + 2 πn, x = - 1.92920 \dots + 2 πn

Degrees

x = 110.53509 \dots^{\circ} + 36 0^{\circ} n, x = - 110.53509 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) - 5 cos (x) - 1 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- 2 + 2 u^{2} - 5 u = 0 : u = \frac{5 + 41}{4}, u = \frac{5 - 41}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 5, c = - 2

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

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

(- 5)^{2} - 4 \cdot 2 (- 2) = 41

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} + 16

5^{2} = 25

= 25 + 16

Add the numbers:

25 + 16 = 41

= 41

u_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 2}

u = \frac{- ( - 5 ) + 41}{2 \cdot 2} : \frac{5 + 41}{4}

\frac{- ( - 5 ) + 41}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 + 41}{4}

u = \frac{- ( - 5 ) - 41}{2 \cdot 2} : \frac{5 - 41}{4}

\frac{- ( - 5 ) - 41}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 - 41}{4}

The solutions to the quadratic equation are:

u = \frac{5 + 41}{4}, u = \frac{5 - 41}{4}

Substitute back

u = cos (x)

cos (x) = \frac{5 + 41}{4}, cos (x) = \frac{5 - 41}{4}

cos (x) = \frac{5 + 41}{4}, cos (x) = \frac{5 - 41}{4}

cos (x) = \frac{5 + 41}{4} :

No Solution

cos (x) = \frac{5 + 41}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{5 - 41}{4} : x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

cos (x) = \frac{5 - 41}{4}

Apply trig inverse properties

cos (x) = \frac{5 - 41}{4}

General solutions for

cos (x) = \frac{5 - 41}{4}

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

x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

Show solutions in decimal form

x = 1.92920 \dots + 2 πn, x = - 1.92920 \dots + 2 πn

Popular Examples

3+3cos(θ)=3

3 + 3 cos (θ) = 3

csc^2(x)=3csc(x)+4

csc^{2} (x) = 3 csc (x) + 4

4cot(2θ)=0

4 cot (2 θ) = 0

sin(60)=cos(x+10)

sin (6 0^{\circ}) = cos (x + 1 0^{\circ})

sin(2x)=682

sin (2 x) = 682

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-5cos(x)-1=0 ?
The general solution for cos(2x)-5cos(x)-1=0 is x=1.92920…+2pin,x=-1.92920…+2pin