What is the general solution for -sin(x)-sin(x+pi/3)=0 ?

The general solution for -sin(x)-sin(x+pi/3)=0 is x=(5pi)/6+pin

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Popular >

-sin(x)-sin(x+pi/3)=0

Solution

- sin (x) - sin (x + \frac{π}{3}) = 0

Solution

x = \frac{5 π}{6} + πn

Degrees

x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

- sin (x) - sin (x + \frac{π}{3}) = 0

Rewrite using trig identities

- sin (x) - sin (x + \frac{π}{3}) = 0

Rewrite using trig identities

sin (x + \frac{π}{3})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (x) cos (\frac{π}{3}) + cos (x) sin (\frac{π}{3})

Simplify

sin (x) cos (\frac{π}{3}) + cos (x) sin (\frac{π}{3}) : \frac{1}{2} sin (x) + \frac{3}{2} cos (x)

sin (x) cos (\frac{π}{3}) + cos (x) sin (\frac{π}{3})

Simplify

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \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}

= \frac{1}{2}

= \frac{1}{2} sin (x) + sin (\frac{π}{3}) cos (x)

Simplify

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= \frac{1}{2} sin (x) + \frac{3}{2} cos (x)

= \frac{1}{2} sin (x) + \frac{3}{2} cos (x)

- sin (x) - (\frac{1}{2} sin (x) + \frac{3}{2} cos (x)) = 0

Simplify

- sin (x) - (\frac{1}{2} sin (x) + \frac{3}{2} cos (x)) : - \frac{3}{2} sin (x) - \frac{3}{2} cos (x)

- sin (x) - (\frac{1}{2} sin (x) + \frac{3}{2} cos (x))

- (\frac{1}{2} sin (x) + \frac{3}{2} cos (x)) : - \frac{1}{2} sin (x) - \frac{3}{2} cos (x)

- (\frac{1}{2} sin (x) + \frac{3}{2} cos (x))

Distribute parentheses

= - (\frac{1}{2} sin (x)) - (\frac{3}{2} cos (x))

Apply minus-plus rules

+ (- a) = - a

= - \frac{1}{2} sin (x) - \frac{3}{2} cos (x)

= - sin (x) - \frac{1}{2} sin (x) - \frac{3}{2} cos (x)

Add similar elements:

- sin (x) - \frac{1}{2} sin (x) = - \frac{3}{2} sin (x)

- sin (x) - \frac{1}{2} sin (x)

Factor out common term

sin (x)

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

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

- 1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- 1 \cdot 2 - 1 = - 3

- 1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= - 2 - 1

Subtract the numbers:

- 2 - 1 = - 3

= - 3

= \frac{- 3}{2}

Apply the fraction rule:

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

= - \frac{3}{2}

= - \frac{3}{2} sin (x)

= - \frac{3}{2} sin (x) - \frac{3}{2} cos (x)

- \frac{3}{2} sin (x) - \frac{3}{2} cos (x) = 0

- \frac{3}{2} sin (x) - \frac{3}{2} cos (x) = 0

Simplify

- \frac{3}{2} sin (x) - \frac{3}{2} cos (x) : \frac{- 3 sin ( x ) - 3 cos ( x )}{2}

- \frac{3}{2} sin (x) - \frac{3}{2} cos (x)

Multiply

\frac{3}{2} sin (x) : \frac{3 sin ( x )}{2}

\frac{3}{2} sin (x)

Multiply fractions:

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

= \frac{3 sin ( x )}{2}

= - \frac{3 sin ( x )}{2} - \frac{3}{2} cos (x)

Multiply

\frac{3}{2} cos (x) : \frac{3 cos ( x )}{2}

\frac{3}{2} cos (x)

Multiply fractions:

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

= \frac{3 cos ( x )}{2}

= - \frac{3 sin ( x )}{2} - \frac{3 cos ( x )}{2}

Apply rule

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

= \frac{- 3 sin ( x ) - 3 cos ( x )}{2}

\frac{- 3 sin ( x ) - 3 cos ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 3 sin (x) - 3 cos (x) = 0

Rewrite using trig identities

- 3 sin (x) - 3 cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{- 3 sin ( x ) - 3 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

- \frac{3 sin ( x )}{cos ( x )} - 3 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 3 tan (x) - 3 = 0

- 3 tan (x) - 3 = 0

Move

3

to the right side

- 3 tan (x) - 3 = 0

Add

3

to both sides

- 3 tan (x) - 3 + 3 = 0 + 3

Simplify

- 3 tan (x) = 3

- 3 tan (x) = 3

Divide both sides by

- 3

- 3 tan (x) = 3

Divide both sides by

- 3

\frac{- 3 tan ( x )}{- 3} = \frac{3}{- 3}

Simplify

tan (x) = - \frac{3}{3}

tan (x) = - \frac{3}{3}

General solutions for

tan (x) = - \frac{3}{3}

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -sin(x)-sin(x+pi/3)=0 ?
The general solution for -sin(x)-sin(x+pi/3)=0 is x=(5pi)/6+pin