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Popular >

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Factor

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

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

Rewrite as

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

Factor out common term

2 cos (x)

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

Factor

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

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

Factor

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

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

Rewrite

2 cos^{2} (x) - 1

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

2 cos^{2} (x) - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

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

Solving each part separately

cos (x) = 0 or cos (2 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (2 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1) = 0 : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

Rewrite using trig identities

cos (2 x) - (- 1 + cos (x) 2) (1 + cos (x) 2) \cdot 2

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

Expand

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

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

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = 2 cos (x), b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

(2 cos (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 2 cos^{2} (x)

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 2 \cdot 2 cos^{2} (x) + 2 \cdot 1 : - 4 cos^{2} (x) + 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

1 - 2 u^{2} = 0

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

1 - 2 u^{2} = 0

Move

1

to the right side

1 - 2 u^{2} = 0

Subtract

1

from both sides

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

Simplify

- 2 u^{2} = - 1

- 2 u^{2} = - 1

Divide both sides by

- 2

- 2 u^{2} = - 1

Divide both sides by

- 2

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

Simplify

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Substitute back

u = cos (x)

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

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

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

cos (x) = \frac{1}{2}

Apply trig inverse properties

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Combine all the solutions

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

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

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