What is the general solution for 4cos^2(x)-5cos(x)+1=0 ?

The general solution for 4cos^2(x)-5cos(x)+1=0 is x=2pin,x=1.31811…+2pin,x=2pi-1.31811…+2pin

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4cos^2(x)-5cos(x)+1=0

Solution

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

Solution

x = 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

4 u^{2} - 5 u + 1 = 0 : u = 1, u = \frac{1}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 5, c = 1

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

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

(- 5)^{2} - 4 \cdot 4 \cdot 1 = 3

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

u = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

\frac{- ( - 5 ) + 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 + 3}{2 \cdot 4}

Add the numbers:

5 + 3 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 5 ) - 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 - 3}{2 \cdot 4}

Subtract the numbers:

5 - 3 = 2

= \frac{2}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{4}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{1}{4}

cos (x) = 1, cos (x) = \frac{1}{4}

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}{4} : x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = \frac{1}{4}

Apply trig inverse properties

cos (x) = \frac{1}{4}

General solutions for

cos (x) = \frac{1}{4}

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

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-5cos(x)+1=0 ?
The general solution for 4cos^2(x)-5cos(x)+1=0 is x=2pin,x=1.31811…+2pin,x=2pi-1.31811…+2pin