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Popular Trigonometry >

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

Solution

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

Solution

x = 2 πn, x = 1.71777 \dots + 2 πn, x = - 1.71777 \dots + 2 πn, x = 2.59356 \dots + 2 πn, x = - 2.59356 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 98.42105 \dots^{\circ} + 36 0^{\circ} n, x = - 98.42105 \dots^{\circ} + 36 0^{\circ} n, x = 148.60028 \dots^{\circ} + 36 0^{\circ} n, x = - 148.60028 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 + cos (x)

from both sides

2 cos (3 x) - 1 - cos (x) = 0

Rewrite using trig identities

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

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

Simplify

- 1 - cos (x) + 2 (4 cos^{3} (x) - 3 cos (x)) : - 7 cos (x) + 8 cos^{3} (x) - 1

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

Expand

2 (4 cos^{3} (x) - 3 cos (x)) : 8 cos^{3} (x) - 6 cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 4 cos^{3} (x), c = 3 cos (x)

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

Simplify

2 \cdot 4 cos^{3} (x) - 2 \cdot 3 cos (x) : 8 cos^{3} (x) - 6 cos (x)

2 \cdot 4 cos^{3} (x) - 2 \cdot 3 cos (x)

Multiply the numbers:

2 \cdot 4 = 8

= 8 cos^{3} (x) - 2 \cdot 3 cos (x)

Multiply the numbers:

2 \cdot 3 = 6

= 8 cos^{3} (x) - 6 cos (x)

= 8 cos^{3} (x) - 6 cos (x)

= - 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x)

Simplify

- 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x) : - 7 cos (x) + 8 cos^{3} (x) - 1

- 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x)

Group like terms

= - cos (x) + 8 cos^{3} (x) - 6 cos (x) - 1

Add similar elements:

- cos (x) - 6 cos (x) = - 7 cos (x)

= - 7 cos (x) + 8 cos^{3} (x) - 1

= - 7 cos (x) + 8 cos^{3} (x) - 1

= - 7 cos (x) + 8 cos^{3} (x) - 1

- 1 - 7 cos (x) + 8 cos^{3} (x) = 0

Solve by substitution

- 1 - 7 cos (x) + 8 cos^{3} (x) = 0

Let:

cos (x) = u

- 1 - 7 u + 8 u^{3} = 0

- 1 - 7 u + 8 u^{3} = 0 : u = 1, u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

- 1 - 7 u + 8 u^{3} = 0

Write in the standard form

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

8 u^{3} - 7 u - 1 = 0

Factor

8 u^{3} - 7 u - 1 : (u - 1) (8 u^{2} + 8 u + 1)

8 u^{3} - 7 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 8

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4, 8

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4 , 8}

\frac{1}{1}

is a root of the expression, so factor out

u - 1

= (u - 1) \frac{8 u ^{3} - 7 u - 1}{u - 1}

\frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + 8 u + 1

\frac{8 u ^{3} - 7 u - 1}{u - 1}

Divide

\frac{8 u ^{3} - 7 u - 1}{u - 1} : \frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

Divide the leading coefficients of the numerator

8 u^{3} - 7 u - 1

and the divisor

u - 1 : \frac{8 u ^{3}}{u} = 8 u^{2}

Quotient = 8 u^{2}

Multiply

u - 1

8 u^{2} : 8 u^{3} - 8 u^{2}

Subtract

8 u^{3} - 8 u^{2}

from

8 u^{3} - 7 u - 1

to get new remainder

Remainder = 8 u^{2} - 7 u - 1

Therefore

\frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

= 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

Divide

\frac{8 u ^{2} - 7 u - 1}{u - 1} : \frac{8 u ^{2} - 7 u - 1}{u - 1} = 8 u + \frac{u - 1}{u - 1}

Divide the leading coefficients of the numerator

8 u^{2} - 7 u - 1

and the divisor

u - 1 : \frac{8 u ^{2}}{u} = 8 u

Quotient = 8 u

Multiply

u - 1

8 u : 8 u^{2} - 8 u

Subtract

8 u^{2} - 8 u

from

8 u^{2} - 7 u - 1

to get new remainder

Remainder = u - 1

Therefore

\frac{8 u ^{2} - 7 u - 1}{u - 1} = 8 u + \frac{u - 1}{u - 1}

= 8 u^{2} + 8 u + \frac{u - 1}{u - 1}

Divide

\frac{u - 1}{u - 1} : \frac{u - 1}{u - 1} = 1

Divide the leading coefficients of the numerator

u - 1

and the divisor

u - 1 : \frac{u}{u} = 1

Quotient = 1

Multiply

u - 1

1 : u - 1

Subtract

u - 1

from

u - 1

to get new remainder

Remainder = 0

Therefore

\frac{u - 1}{u - 1} = 1

= 8 u^{2} + 8 u + 1

= (u - 1) (8 u^{2} + 8 u + 1)

(u - 1) (8 u^{2} + 8 u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u - 1 = 0 or 8 u^{2} + 8 u + 1 = 0

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

8 u^{2} + 8 u + 1 = 0 : u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

8 u^{2} + 8 u + 1 = 0

Solve with the quadratic formula

8 u^{2} + 8 u + 1 = 0

Quadratic Equation Formula:

For

a = 8, b = 8, c = 1

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

8^{2} - 4 \cdot 8 \cdot 1 = 42

8^{2} - 4 \cdot 8 \cdot 1

Multiply the numbers:

4 \cdot 8 \cdot 1 = 32

= 8^{2} - 32

8^{2} = 64

= 64 - 32

Subtract the numbers:

64 - 32 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

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

= 2^{4} \cdot 2

Apply radical rule:

n ab = n a n b

= 2 2^{4}

Apply radical rule:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

u_{1, 2} = \frac{- 8 \pm 4 2}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 8 + 4 2}{2 \cdot 8}, u_{2} = \frac{- 8 - 4 2}{2 \cdot 8}

u = \frac{- 8 + 4 2}{2 \cdot 8} : \frac{- 2 + 2}{4}

\frac{- 8 + 4 2}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8 + 4 2}{16}

Factor

- 8 + 42 : 4 (- 2 + 2)

- 8 + 42

Rewrite as

= - 4 \cdot 2 + 42

Factor out common term

4

= 4 (- 2 + 2)

= \frac{4 ( - 2 + 2 )}{16}

Cancel the common factor:

4

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

u = \frac{- 8 - 4 2}{2 \cdot 8} : - \frac{2 + 2}{4}

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8 - 4 2}{16}

Factor

- 8 - 42 : - 4 (2 + 2)

- 8 - 42

Rewrite as

= - 4 \cdot 2 - 42

Factor out common term

4

= - 4 (2 + 2)

= - \frac{4 ( 2 + 2 )}{16}

Cancel the common factor:

4

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

The solutions to the quadratic equation are:

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

The solutions are

u = 1, u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{- 2 + 2}{4}, cos (x) = - \frac{2 + 2}{4}

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

cos (x) = \frac{- 2 + 2}{4}

Apply trig inverse properties

cos (x) = \frac{- 2 + 2}{4}

General solutions for

cos (x) = \frac{- 2 + 2}{4}

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

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

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

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

cos (x) = - \frac{2 + 2}{4}

Apply trig inverse properties

cos (x) = - \frac{2 + 2}{4}

General solutions for

cos (x) = - \frac{2 + 2}{4}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 2 πn, x = 1.71777 \dots + 2 πn, x = - 1.71777 \dots + 2 πn, x = 2.59356 \dots + 2 πn, x = - 2.59356 \dots + 2 πn

Graph

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