What is the general solution for cos(x+x)+cos(x)=0 ?

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

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cos(x+x)+cos(x)=0

Solution

cos (x + x) + cos (x) = 0

Solution

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

Degrees

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

Solution steps

cos (x + x) + cos (x) = 0

Rewrite using trig identities

cos (x + x) + cos (x) = 0

Rewrite using trig identities

cos (x + x)

Use the Angle Sum identity:

cos (s + s) = cos (s) cos (s) - sin (s) sin (s)

= cos (x) cos (x) - sin (x) sin (x)

Simplify

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

cos (x) cos (x) - sin (x) sin (x)

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

sin (x) sin (x)

Apply exponent rule:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

cos^{2} (x) - sin^{2} (x) + cos (x) = 0

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 1

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

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

1^{2} - 4 \cdot 2 (- 1) = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

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

Combine all the solutions

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

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sin (x + 3 0^{\circ}) = cos (x + 6 0^{\circ})

cos(5x+40)= 3/5

cos (5 x + 4 0^{\circ}) = \frac{3}{5}

cos(4x)+cos(2x)+1=0

cos (4 x) + cos (2 x) + 1 = 0

sin(2x)=0.3,-180\circ <= x<= 180

sin (2 x) = 0.3, - 180 \circ \leq x \leq 18 0^{\circ}

2^{cot^2(x)}=8

2^{c o t^{2} (x)} = 8

Frequently Asked Questions (FAQ)

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