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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 18)^{2} - 4 \cdot 9 \cdot 9 = 0

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

Apply exponent rule:

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

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} - 4 \cdot 9 \cdot 9

Multiply the numbers:

4 \cdot 9 \cdot 9 = 324

= 1 8^{2} - 324

1 8^{2} = 324

= 324 - 324

Subtract the numbers:

324 - 324 = 0

= 0

u_{1, 2} = \frac{- ( - 18 ) \pm 0}{2 \cdot 9}

u = \frac{- ( - 18 )}{2 \cdot 9}

\frac{- ( - 18 )}{2 \cdot 9} = 1

\frac{- ( - 18 )}{2 \cdot 9}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

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

Combine all the solutions

x = 2 πn

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

sin (a) = - \frac{5}{13}

sin^{2} (x) = \frac{( 8 m - 2 )}{9}

0.06 = 0.1 (cos (4 (t) - 1.57))

sin (x) = \frac{13}{14}

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