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

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

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

Solution

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

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

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

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

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- 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 the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Remove parentheses:

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