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Popular >

8cos(2x)+6=cos^2(x)+cos(x)

Solution

8 cos (2 x) + 6 = cos^{2} (x) + cos (x)

Solution

x = 1.15927 \dots + 2 πn, x = 2 π - 1.15927 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

Degrees

x = 66.42182 \dots^{\circ} + 36 0^{\circ} n, x = 293.57817 \dots^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 cos (2 x) + 6 = cos^{2} (x) + cos (x)

Subtract

cos^{2} (x) + cos (x)

from both sides

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

Rewrite using trig identities

6 - cos (x) - cos^{2} (x) + 8 cos (2 x)

Use the Double Angle identity:

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

= 6 - cos (x) - cos^{2} (x) + 8 (2 cos^{2} (x) - 1)

Simplify

6 - cos (x) - cos^{2} (x) + 8 (2 cos^{2} (x) - 1) : 15 cos^{2} (x) - cos (x) - 2

6 - cos (x) - cos^{2} (x) + 8 (2 cos^{2} (x) - 1)

Expand

8 (2 cos^{2} (x) - 1) : 16 cos^{2} (x) - 8

8 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 2 cos^{2} (x), c = 1

= 8 \cdot 2 cos^{2} (x) - 8 \cdot 1

Simplify

8 \cdot 2 cos^{2} (x) - 8 \cdot 1 : 16 cos^{2} (x) - 8

8 \cdot 2 cos^{2} (x) - 8 \cdot 1

Multiply the numbers:

8 \cdot 2 = 16

= 16 cos^{2} (x) - 8 \cdot 1

Multiply the numbers:

8 \cdot 1 = 8

= 16 cos^{2} (x) - 8

= 16 cos^{2} (x) - 8

= 6 - cos (x) - cos^{2} (x) + 16 cos^{2} (x) - 8

Simplify

6 - cos (x) - cos^{2} (x) + 16 cos^{2} (x) - 8 : 15 cos^{2} (x) - cos (x) - 2

6 - cos (x) - cos^{2} (x) + 16 cos^{2} (x) - 8

Add similar elements:

- cos^{2} (x) + 16 cos^{2} (x) = 15 cos^{2} (x)

= 6 - cos (x) + 15 cos^{2} (x) - 8

Group like terms

= - cos (x) + 15 cos^{2} (x) + 6 - 8

Add/Subtract the numbers:

6 - 8 = - 2

= 15 cos^{2} (x) - cos (x) - 2

= 15 cos^{2} (x) - cos (x) - 2

= 15 cos^{2} (x) - cos (x) - 2

- 2 - cos (x) + 15 cos^{2} (x) = 0

Solve by substitution

- 2 - cos (x) + 15 cos^{2} (x) = 0

Let:

cos (x) = u

- 2 - u + 15 u^{2} = 0

- 2 - u + 15 u^{2} = 0 : u = \frac{2}{5}, u = - \frac{1}{3}

- 2 - u + 15 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

15 u^{2} - u - 2 = 0

Solve with the quadratic formula

15 u^{2} - u - 2 = 0

Quadratic Equation Formula:

For

a = 15, b = - 1, c = - 2

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 15 ( - 2 )}{2 \cdot 15}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 15 ( - 2 )}{2 \cdot 15}

(- 1)^{2} - 4 \cdot 15 (- 2) = 11

(- 1)^{2} - 4 \cdot 15 (- 2)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 15 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 15 \cdot 2 = 120

4 \cdot 15 \cdot 2

Multiply the numbers:

4 \cdot 15 \cdot 2 = 120

= 120

= 1 + 120

Add the numbers:

1 + 120 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 15}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 15}, u_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 15}

u = \frac{- ( - 1 ) + 11}{2 \cdot 15} : \frac{2}{5}

\frac{- ( - 1 ) + 11}{2 \cdot 15}

Apply rule

- (- a) = a

= \frac{1 + 11}{2 \cdot 15}

Add the numbers:

1 + 11 = 12

= \frac{12}{2 \cdot 15}

Multiply the numbers:

2 \cdot 15 = 30

= \frac{12}{30}

Cancel the common factor:

6

= \frac{2}{5}

u = \frac{- ( - 1 ) - 11}{2 \cdot 15} : - \frac{1}{3}

\frac{- ( - 1 ) - 11}{2 \cdot 15}

Apply rule

- (- a) = a

= \frac{1 - 11}{2 \cdot 15}

Subtract the numbers:

1 - 11 = - 10

= \frac{- 10}{2 \cdot 15}

Multiply the numbers:

2 \cdot 15 = 30

= \frac{- 10}{30}

Apply the fraction rule:

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

= - \frac{10}{30}

Cancel the common factor:

10

= - \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{2}{5}, u = - \frac{1}{3}

Substitute back

u = cos (x)

cos (x) = \frac{2}{5}, cos (x) = - \frac{1}{3}

cos (x) = \frac{2}{5}, cos (x) = - \frac{1}{3}

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

cos (x) = \frac{2}{5}

Apply trig inverse properties

cos (x) = \frac{2}{5}

General solutions for

cos (x) = \frac{2}{5}

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

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

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

cos (x) = - \frac{1}{3} : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

cos (x) = - \frac{1}{3}

Apply trig inverse properties

cos (x) = - \frac{1}{3}

General solutions for

cos (x) = - \frac{1}{3}

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

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{2}{5}) + 2 πn, x = 2 π - arccos (\frac{2}{5}) + 2 πn, x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 1.15927 \dots + 2 πn, x = 2 π - 1.15927 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

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