What is the general solution for 9cos^2(x)-9cos(x)-18=0 ?

The general solution for 9cos^2(x)-9cos(x)-18=0 is x=pi+2pin

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9cos^2(x)-9cos(x)-18=0

Solution

9 cos^{2} (x) - 9 cos (x) - 18 = 0

Solution

x = π + 2 πn

Degrees

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

Solution steps

9 cos^{2} (x) - 9 cos (x) - 18 = 0

Solve by substitution

9 cos^{2} (x) - 9 cos (x) - 18 = 0

Let:

cos (x) = u

9 u^{2} - 9 u - 18 = 0

9 u^{2} - 9 u - 18 = 0 : u = 2, u = - 1

9 u^{2} - 9 u - 18 = 0

Solve with the quadratic formula

9 u^{2} - 9 u - 18 = 0

Quadratic Equation Formula:

For

a = 9, b = - 9, c = - 18

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 9 ( - 18 )}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 9 ( - 18 )}{2 \cdot 9}

(- 9)^{2} - 4 \cdot 9 (- 18) = 27

(- 9)^{2} - 4 \cdot 9 (- 18)

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 9 \cdot 18

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 9 \cdot 18

Multiply the numbers:

4 \cdot 9 \cdot 18 = 648

= 9^{2} + 648

9^{2} = 81

= 81 + 648

Add the numbers:

81 + 648 = 729

= 729

Factor the number:

729 = 2 7^{2}

= 2 7^{2}

Apply radical rule:

n a^{n} = a

2 7^{2} = 27

= 27

u_{1, 2} = \frac{- ( - 9 ) \pm 27}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 27}{2 \cdot 9}, u_{2} = \frac{- ( - 9 ) - 27}{2 \cdot 9}

u = \frac{- ( - 9 ) + 27}{2 \cdot 9} : 2

\frac{- ( - 9 ) + 27}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{9 + 27}{2 \cdot 9}

Add the numbers:

9 + 27 = 36

= \frac{36}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{36}{18}

Divide the numbers:

\frac{36}{18} = 2

= 2

u = \frac{- ( - 9 ) - 27}{2 \cdot 9} : - 1

\frac{- ( - 9 ) - 27}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{9 - 27}{2 \cdot 9}

Subtract the numbers:

9 - 27 = - 18

= \frac{- 18}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 18}{18}

Apply the fraction rule:

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

= - \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 2, u = - 1

Substitute back

u = cos (x)

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

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

cos (x) = 2 :

No Solution

cos (x) = 2

- 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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 9cos^2(x)-9cos(x)-18=0 ?
The general solution for 9cos^2(x)-9cos(x)-18=0 is x=pi+2pin