What is the general solution for tan(x+pi)+2sin(x+pi)=0,0<= x<= 2pi ?

Q: What is the general solution for tan(x+pi)+2sin(x+pi)=0,0<= x<= 2pi ?

The general solution for tan(x+pi)+2sin(x+pi)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x= pi/3 ,x=(5pi)/3

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tan(x+pi)+2sin(x+pi)=0,0<= x<= 2pi

Solution

tan (x + π) + 2 sin (x + π) = 0, 0 \leq x \leq 2 π

Solution

x = 0, x = π, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

Degrees

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

Solution steps

tan (x + π) + 2 sin (x + π) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

tan (x + π) + 2 sin (x + π) = 0

Rewrite using trig identities

tan (x + π)

Use the basic trigonometric identity:

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

= \frac{sin ( x + π )}{cos ( x + π )}

Use the Angle Sum identity:

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

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

Use the Angle Sum identity:

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

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

Simplify

\frac{sin ( x ) cos ( π ) + cos ( x ) sin ( π )}{cos ( x ) cos ( π ) - sin ( x ) sin ( π )} : \frac{sin ( x )}{cos ( x )}

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

sin (x) cos (π) + cos (x) sin (π) = - sin (x)

sin (x) cos (π) + cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 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}

= - 1

= (- 1) sin (x)

Refine

= - sin (x)

= - sin (x) + sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

= - sin (x) + 0

- sin (x) + 0 = - sin (x)

= - sin (x)

= \frac{- sin ( x )}{cos ( π ) cos ( x ) - sin ( π ) sin ( x )}

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

cos (x) cos (π) - sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 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}

= - 1

= (- 1) cos (x)

Refine

= - cos (x)

= - cos (x) - sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

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}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) - 0

- cos (x) - 0 = - cos (x)

= - cos (x)

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

Apply the fraction rule:

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

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

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

Use the Angle Sum identity:

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

= sin (x) cos (π) + cos (x) sin (π)

Simplify

sin (x) cos (π) + cos (x) sin (π) : - sin (x)

sin (x) cos (π) + cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 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}

= - 1

= (- 1) sin (x)

Refine

= - sin (x)

= - sin (x) + sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

= - sin (x) + 0

- sin (x) + 0 = - sin (x)

= - sin (x)

= - sin (x)

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Factor

sin (x) - 2 sin (x) cos (x) : - sin (x) (2 cos (x) - 1)

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

Factor out common term

- sin (x)

= - sin (x) (- 1 + 2 cos (x))

- sin (x) (2 cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) - 1 = 0

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

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

Simplify

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

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sin (5 x) = 1

Frequently Asked Questions (FAQ)

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