What is the general solution for cos(2x)*2-2cos(x)=0 ?

The general solution for cos(2x)*2-2cos(x)=0 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2x)*2-2cos(x)=0

Solution

cos (2 x) \cdot 2 - 2 cos (x) = 0

Solution

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) \cdot 2 - 2 cos (x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

(- 1 + 2 u^{2}) \cdot 2 - 2 u = 0 : u = 1, u = - \frac{1}{2}

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

Expand

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

(- 1 + 2 u^{2}) \cdot 2 - 2 u

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

Expand

2 (- 1 + 2 u^{2}) : - 2 + 4 u^{2}

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

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 1, c = 2 u^{2}

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

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 1 + 2 \cdot 2 u^{2}

Simplify

- 2 \cdot 1 + 2 \cdot 2 u^{2} : - 2 + 4 u^{2}

- 2 \cdot 1 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 2 = 4

= - 2 + 4 u^{2}

= - 2 + 4 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 \cdot 4 (- 2) = 6

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 4}

u = \frac{- ( - 2 ) + 6}{2 \cdot 4} : 1

\frac{- ( - 2 ) + 6}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 + 6}{2 \cdot 4}

Add the numbers:

2 + 6 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 6}{2 \cdot 4} : - \frac{1}{2}

\frac{- ( - 2 ) - 6}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 - 6}{2 \cdot 4}

Subtract the numbers:

2 - 6 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

-2sin(2x)= pi/(12)

- 2 sin (2 x) = \frac{π}{12}

cos(x)=(sec(x))/4

cos (x) = \frac{sec ( x )}{4}

sin(10x)=sin(10(x+1))

sin (10 x) = sin (10 (x + 1))

sin^2(θ)+3sin(θ)=0

sin^{2} (θ) + 3 sin (θ) = 0

cos(x)=-0.7962

cos (x) = - 0.7962

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)*2-2cos(x)=0 ?
The general solution for cos(2x)*2-2cos(x)=0 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin