What is the general solution for cos^2(x)+5cos(x)-6=0 ?

The general solution for cos^2(x)+5cos(x)-6=0 is x=2pin

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cos^2(x)+5cos(x)-6=0

Solution

cos^{2} (x) + 5 cos (x) - 6 = 0

Solution

x = 2 πn

Degrees

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

Solution steps

cos^{2} (x) + 5 cos (x) - 6 = 0

Solve by substitution

cos^{2} (x) + 5 cos (x) - 6 = 0

Let:

cos (x) = u

u^{2} + 5 u - 6 = 0

u^{2} + 5 u - 6 = 0 : u = 1, u = - 6

u^{2} + 5 u - 6 = 0

Solve with the quadratic formula

u^{2} + 5 u - 6 = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 6) = 7

5^{2} - 4 \cdot 1 \cdot (- 6)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 \cdot 1}, u_{2} = \frac{- 5 - 7}{2 \cdot 1}

u = \frac{- 5 + 7}{2 \cdot 1} : 1

\frac{- 5 + 7}{2 \cdot 1}

Add/Subtract the numbers:

- 5 + 7 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 5 - 7}{2 \cdot 1} : - 6

\frac{- 5 - 7}{2 \cdot 1}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 12}{2}

Apply the fraction rule:

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

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

The solutions to the quadratic equation are:

u = 1, u = - 6

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 6

cos (x) = 1, cos (x) = - 6

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) = - 6 :

No Solution

cos (x) = - 6

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

10sin(x)=4cos^2(x)+2

10 sin (x) = 4 cos^{2} (x) + 2

-4+sin(-3x+(2pi)/3)=-4

- 4 + sin (- 3 x + \frac{2 π}{3}) = - 4

2*sin(x)*cos(x)=cos(x)

2 \cdot sin (x) \cdot cos (x) = cos (x)

sin(2θ)=(-2sqrt(6))/7

sin (2 θ) = \frac{- 2 6}{7}

solvefor x,sin(x)=-7/25 solve for

x, sin (x) = - \frac{7}{25}

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)+5cos(x)-6=0 ?
The general solution for cos^2(x)+5cos(x)-6=0 is x=2pin