What is the general solution for solvefor x,-4cos(x)-4cos(2x)=0 ?

The general solution for solvefor x,-4cos(x)-4cos(2x)=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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solvefor x,-4cos(x)-4cos(2x)=0

Solution

solve for

x, - 4 cos (x) - 4 cos (2 x) = 0

Solution

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos (x) - 4 cos (2 x) = 0

Rewrite using trig identities

- 4 cos (2 x) - 4 cos (x)

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- (- 1 + 2 u^{2}) \cdot 4 - 4 u = 0

- (- 1 + 2 u^{2}) \cdot 4 - 4 u = 0 : u = - 1, u = \frac{1}{2}

- (- 1 + 2 u^{2}) \cdot 4 - 4 u = 0

Expand

- (- 1 + 2 u^{2}) \cdot 4 - 4 u : 4 - 8 u^{2} - 4 u

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

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

Expand

- 4 (- 1 + 2 u^{2}) : 4 - 8 u^{2}

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

Apply the distributive law:

a (b + c) = ab + a c

a = - 4, b = - 1, c = 2 u^{2}

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

Apply minus-plus rules

- (- a) = a, + (- a) = - a

= 4 \cdot 1 - 4 \cdot 2 u^{2}

Simplify

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

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

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 2 = 8

= 4 - 8 u^{2}

= 4 - 8 u^{2}

= 4 - 8 u^{2} - 4 u

4 - 8 u^{2} - 4 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 8, b = - 4, c = 4

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

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

(- 4)^{2} - 4 (- 8) \cdot 4 = 12

(- 4)^{2} - 4 (- 8) \cdot 4

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 8 \cdot 4

Apply exponent rule:

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

n

is even

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

= 4^{2} + 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 4^{2} + 128

4^{2} = 16

= 16 + 128

Add the numbers:

16 + 128 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 12}{2 ( - 8 )}, u_{2} = \frac{- ( - 4 ) - 12}{2 ( - 8 )}

u = \frac{- ( - 4 ) + 12}{2 ( - 8 )} : - 1

\frac{- ( - 4 ) + 12}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{4 + 12}{- 2 \cdot 8}

Add the numbers:

4 + 12 = 16

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{16}{- 16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 4 ) - 12}{2 ( - 8 )} : \frac{1}{2}

\frac{- ( - 4 ) - 12}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{4 - 12}{- 2 \cdot 8}

Subtract the numbers:

4 - 12 = - 8

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8}{- 16}

Apply the fraction rule:

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

= \frac{8}{16}

Cancel the common factor:

8

= \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 = π + 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 = π + 2 πn

x = π + 2 πn

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,-4cos(x)-4cos(2x)=0 ?
The general solution for solvefor x,-4cos(x)-4cos(2x)=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin