What is the general solution for cos(3x)-21cos(x)+16=0 ?

The general solution for cos(3x)-21cos(x)+16=0 is x=0.74946…+2pin,x=2pi-0.74946…+2pin

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Popular Trigonometry >

cos(3x)-21cos(x)+16=0

Solution

cos (3 x) - 21 cos (x) + 16 = 0

Solution

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

Degrees

x = 42.94140 \dots^{\circ} + 36 0^{\circ} n, x = 317.05859 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (3 x) - 21 cos (x) + 16 = 0

Rewrite using trig identities

16 + cos (3 x) - 21 cos (x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

Rewrite using trig identities

cos (3 x)

Rewrite as

= cos (2 x + x)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 x) cos (x) - sin (2 x) sin (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

Simplify

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= 2 cos (x) sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Expand

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

Expand

cos (x) (2 cos^{2} (x) - 1) : 2 cos^{3} (x) - cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

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

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x) - 2 (1 - cos^{2} (x)) cos (x)

Expand

- 2 cos (x) (1 - cos^{2} (x)) : - 2 cos (x) + 2 cos^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Simplify

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Group like terms

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 16 + 4 cos^{3} (x) - 3 cos (x) - 21 cos (x)

Simplify

= 16 + 4 cos^{3} (x) - 24 cos (x)

16 - 24 cos (x) + 4 cos^{3} (x) = 0

Solve by substitution

16 - 24 cos (x) + 4 cos^{3} (x) = 0

Let:

cos (x) = u

16 - 24 u + 4 u^{3} = 0

16 - 24 u + 4 u^{3} = 0 : u = 2, u = - 1 + 3, u = - 1 - 3

16 - 24 u + 4 u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 u^{3} - 24 u + 16 = 0

Factor

4 u^{3} - 24 u + 16 : 4 (u - 2) (u^{2} + 2 u - 2)

4 u^{3} - 24 u + 16

Factor out common term

4 : 4 (u^{3} - 6 u + 4)

4 u^{3} - 24 u + 16

Rewrite

16

4 \cdot 4

Rewrite

24

4 \cdot 6

= 4 u^{3} - 4 \cdot 6 u + 4 \cdot 4

Factor out common term

4

= 4 (u^{3} - 6 u + 4)

= 4 (u^{3} - 6 u + 4)

Factor

u^{3} - 6 u + 4 : (u - 2) (u^{2} + 2 u - 2)

u^{3} - 6 u + 4

Use the rational root theorem

a_{0} = 4, a_{n} = 1

The dividers of

a_{0} : 1, 2, 4,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 2 , 4}{1}

\frac{2}{1}

is a root of the expression, so factor out

u - 2

= (u - 2) \frac{u ^{3} - 6 u + 4}{u - 2}

\frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + 2 u - 2

\frac{u ^{3} - 6 u + 4}{u - 2}

Divide

\frac{u ^{3} - 6 u + 4}{u - 2} : \frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

Divide the leading coefficients of the numerator

u^{3} - 6 u + 4

and the divisor

u - 2 : \frac{u ^{3}}{u} = u^{2}

Quotient = u^{2}

Multiply

u - 2

u^{2} : u^{3} - 2 u^{2}

Subtract

u^{3} - 2 u^{2}

from

u^{3} - 6 u + 4

to get new remainder

Remainder = 2 u^{2} - 6 u + 4

Therefore

\frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

= u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

Divide

\frac{2 u ^{2} - 6 u + 4}{u - 2} : \frac{2 u ^{2} - 6 u + 4}{u - 2} = 2 u + \frac{- 2 u + 4}{u - 2}

Divide the leading coefficients of the numerator

2 u^{2} - 6 u + 4

and the divisor

u - 2 : \frac{2 u ^{2}}{u} = 2 u

Quotient = 2 u

Multiply

u - 2

2 u : 2 u^{2} - 4 u

Subtract

2 u^{2} - 4 u

from

2 u^{2} - 6 u + 4

to get new remainder

Remainder = - 2 u + 4

Therefore

\frac{2 u ^{2} - 6 u + 4}{u - 2} = 2 u + \frac{- 2 u + 4}{u - 2}

= u^{2} + 2 u + \frac{- 2 u + 4}{u - 2}

Divide

\frac{- 2 u + 4}{u - 2} : \frac{- 2 u + 4}{u - 2} = - 2

Divide the leading coefficients of the numerator

- 2 u + 4

and the divisor

u - 2 : \frac{- 2 u}{u} = - 2

Quotient = - 2

Multiply

u - 2

- 2 : - 2 u + 4

Subtract

- 2 u + 4

from

- 2 u + 4

to get new remainder

Remainder = 0

Therefore

\frac{- 2 u + 4}{u - 2} = - 2

= u^{2} + 2 u - 2

= u^{2} + 2 u - 2

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

= 4 (u - 2) (u^{2} + 2 u - 2)

4 (u - 2) (u^{2} + 2 u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to both sides

u - 2 + 2 = 0 + 2

Simplify

u = 2

u = 2

Solve

u^{2} + 2 u - 2 = 0 : u = - 1 + 3, u = - 1 - 3

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 \cdot 1 \cdot (- 2) = 23

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

u_{1, 2} = \frac{- 2 \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 2 + 2 3}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 3}{2 \cdot 1}

u = \frac{- 2 + 2 3}{2 \cdot 1} : - 1 + 3

\frac{- 2 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 + 2 3}{2}

Factor

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

Rewrite as

= - 2 \cdot 1 + 23

Factor out common term

2

= 2 (- 1 + 3)

= \frac{2 ( - 1 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 3

u = \frac{- 2 - 2 3}{2 \cdot 1} : - 1 - 3

\frac{- 2 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 3}{2}

Factor

- 2 - 23 : - 2 (1 + 3)

- 2 - 23

Rewrite as

= - 2 \cdot 1 - 23

Factor out common term

2

= - 2 (1 + 3)

= - \frac{2 ( 1 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 3)

Negate

- (1 + 3) = - 1 - 3

= - 1 - 3

The solutions to the quadratic equation are:

u = - 1 + 3, u = - 1 - 3

The solutions are

u = 2, u = - 1 + 3, u = - 1 - 3

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = 2, cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 1 + 3 : x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 + 3

Apply trig inverse properties

cos (x) = - 1 + 3

General solutions for

cos (x) = - 1 + 3

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

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 - 3 :

No Solution

cos (x) = - 1 - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

Show solutions in decimal form

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(3x)-21cos(x)+16=0 ?
The general solution for cos(3x)-21cos(x)+16=0 is x=0.74946…+2pin,x=2pi-0.74946…+2pin