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Popular Trigonometry >

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

Solution

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

Solution

x = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn, x = 1.88495 \dots + 2 πn, x = - 1.88495 \dots + 2 πn

Degrees

x = 3 6^{\circ} + 36 0^{\circ} n, x = 32 4^{\circ} + 36 0^{\circ} n, x = 10 8^{\circ} + 36 0^{\circ} n, x = - 10 8^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 5}{8}

Factor

2 + 25 : 2 (1 + 5)

2 + 25

Rewrite as

= 2 \cdot 1 + 25

Factor out common term

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

Cancel the common factor:

2

= \frac{1 + 5}{4}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 5}{8}

Factor

2 - 25 : 2 (1 - 5)

2 - 25

Rewrite as

= 2 \cdot 1 - 25

Factor out common term

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{8}

Cancel the common factor:

2

= \frac{1 - 5}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 5}{4}, cos (x) = \frac{1 - 5}{4}

cos (x) = \frac{1 + 5}{4}, cos (x) = \frac{1 - 5}{4}

cos (x) = \frac{1 + 5}{4} : x = arccos (\frac{1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

cos (x) = \frac{1 + 5}{4}

Apply trig inverse properties

cos (x) = \frac{1 + 5}{4}

General solutions for

cos (x) = \frac{1 + 5}{4}

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

x = arccos (\frac{1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

x = arccos (\frac{1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

cos (x) = \frac{1 - 5}{4} : x = arccos (\frac{1 - 5}{4}) + 2 πn, x = - arccos (\frac{1 - 5}{4}) + 2 πn

cos (x) = \frac{1 - 5}{4}

Apply trig inverse properties

cos (x) = \frac{1 - 5}{4}

General solutions for

cos (x) = \frac{1 - 5}{4}

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

x = arccos (\frac{1 - 5}{4}) + 2 πn, x = - arccos (\frac{1 - 5}{4}) + 2 πn

x = arccos (\frac{1 - 5}{4}) + 2 πn, x = - arccos (\frac{1 - 5}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn, x = arccos (\frac{1 - 5}{4}) + 2 πn, x = - arccos (\frac{1 - 5}{4}) + 2 πn

Show solutions in decimal form

x = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn, x = 1.88495 \dots + 2 πn, x = - 1.88495 \dots + 2 πn

Graph

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