What is the general solution for cos^2(x)-6cos(x)-7=0 ?

The general solution for cos^2(x)-6cos(x)-7=0 is x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)-6cos(x)-7=0

Solution

cos^{2} (x) - 6 cos (x) - 7 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 6 cos (x) - 7 = 0

Solve by substitution

cos^{2} (x) - 6 cos (x) - 7 = 0

Let:

cos (x) = u

u^{2} - 6 u - 7 = 0

u^{2} - 6 u - 7 = 0 : u = 7, u = - 1

u^{2} - 6 u - 7 = 0

Solve with the quadratic formula

u^{2} - 6 u - 7 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = - 7

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 7) = 8

(- 6)^{2} - 4 \cdot 1 \cdot (- 7)

Apply rule

- (- a) = a

= (- 6)^{2} + 4 \cdot 1 \cdot 7

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} + 4 \cdot 1 \cdot 7

Multiply the numbers:

4 \cdot 1 \cdot 7 = 28

= 6^{2} + 28

6^{2} = 36

= 36 + 28

Add the numbers:

36 + 28 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 6 ) \pm 8}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 8}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 8}{2 \cdot 1}

u = \frac{- ( - 6 ) + 8}{2 \cdot 1} : 7

\frac{- ( - 6 ) + 8}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 8}{2 \cdot 1}

Add the numbers:

6 + 8 = 14

= \frac{14}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{14}{2}

Divide the numbers:

\frac{14}{2} = 7

= 7

u = \frac{- ( - 6 ) - 8}{2 \cdot 1} : - 1

\frac{- ( - 6 ) - 8}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 8}{2 \cdot 1}

Subtract the numbers:

6 - 8 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 7, u = - 1

Substitute back

u = cos (x)

cos (x) = 7, cos (x) = - 1

cos (x) = 7, cos (x) = - 1

cos (x) = 7 :

No Solution

cos (x) = 7

- 1 \leq cos (x) \leq 1

No Solution

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 = π + 2 πn

Popular Examples

cos(θ)=0,0<= θ<= 2pi

cos (θ) = 0, 0 \leq θ \leq 2 π

-2sin(x)-2+4sin^2(x)=0

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

cos(x)=-3/5 ,pi<x<(3pi)/2 ,sin(x/2)

cos (x) = - \frac{3}{5}, π < x < \frac{3 π}{2}, sin (\frac{x}{2})

30=arccos(6/(x(3)))

3 0^{\circ} = arccos (\frac{6}{x ( 3 )})

sin(θ)=0.18421

sin (θ) = 0.18421

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-6cos(x)-7=0 ?
The general solution for cos^2(x)-6cos(x)-7=0 is x=pi+2pin