What is the general solution for 5cos^2(x)-6cos(x)+1=0 ?

The general solution for 5cos^2(x)-6cos(x)+1=0 is x=2pin,x=1.36943…+2pin,x=2pi-1.36943…+2pin

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5cos^2(x)-6cos(x)+1=0

Solution

5 cos^{2} (x) - 6 cos (x) + 1 = 0

Solution

x = 2 πn, x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos^{2} (x) - 6 cos (x) + 1 = 0

Solve by substitution

5 cos^{2} (x) - 6 cos (x) + 1 = 0

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 5 \cdot 1 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- ( - 6 ) + 4}{2 \cdot 5} : 1

\frac{- ( - 6 ) + 4}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{6 + 4}{2 \cdot 5}

Add the numbers:

6 + 4 = 10

= \frac{10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 6 ) - 4}{2 \cdot 5} : \frac{1}{5}

\frac{- ( - 6 ) - 4}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{6 - 4}{2 \cdot 5}

Subtract the numbers:

6 - 4 = 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{5}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{1}{5}

cos (x) = 1, cos (x) = \frac{1}{5}

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) = \frac{1}{5} : x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

cos (x) = \frac{1}{5}

Apply trig inverse properties

cos (x) = \frac{1}{5}

General solutions for

cos (x) = \frac{1}{5}

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

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(x)-6cos(x)+1=0 ?
The general solution for 5cos^2(x)-6cos(x)+1=0 is x=2pin,x=1.36943…+2pin,x=2pi-1.36943…+2pin