What is the general solution for tan(-x)=tan(2x+1) ?

The general solution for tan(-x)=tan(2x+1) is x=(2pin)/3-1/3 ,x= pi/3+(2pin)/3-1/3

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tan(-x)=tan(2x+1)

Solution

tan (- x) = tan (2 x + 1)

Solution

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

Degrees

x = - 19.09859 \dots^{\circ} + 12 0^{\circ} n, x = 40.90140 \dots^{\circ} + 12 0^{\circ} n

Solution steps

tan (- x) = tan (2 x + 1)

Rewrite using trig identities

tan (- x) = tan (2 x + 1)

- tan (x) = tan (2 x + 1)

- tan (x) = tan (2 x + 1)

Subtract

tan (2 x + 1)

from both sides

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

Express with sin, cos

- tan (1 + 2 x) - tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Least Common Multiplier of

cos (1 + 2 x), cos (x) : cos (2 x + 1) cos (x)

cos (1 + 2 x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (1 + 2 x)

cos (x)

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

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (2 x + 1) cos (x)

For

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

multiply the denominator and numerator by

cos (x)

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

For

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

multiply the denominator and numerator by

cos (2 x + 1)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Angle Sum identity:

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

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

= - sin (1 + 2 x + x)

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

Divide both sides by

- 1

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

Divide both sides by

- 1

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

Simplify

sin (1 + 2 x + x) = 0

sin (1 + 2 x + x) = 0

General solutions for

sin (1 + 2 x + 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}

1 + 2 x + x = 0 + 2 πn, 1 + 2 x + x = π + 2 πn

1 + 2 x + x = 0 + 2 πn, 1 + 2 x + x = π + 2 πn

Solve

1 + 2 x + x = 0 + 2 πn : x = \frac{2 πn}{3} - \frac{1}{3}

1 + 2 x + x = 0 + 2 πn

Add similar elements:

2 x + x = 3 x

1 + 3 x = 0 + 2 πn

0 + 2 πn = 2 πn

1 + 3 x = 2 πn

Move

1

to the right side

1 + 3 x = 2 πn

Subtract

1

from both sides

1 + 3 x - 1 = 2 πn - 1

Simplify

3 x = 2 πn - 1

3 x = 2 πn - 1

Divide both sides by

3

3 x = 2 πn - 1

Divide both sides by

3

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

Simplify

x = \frac{2 πn}{3} - \frac{1}{3}

x = \frac{2 πn}{3} - \frac{1}{3}

Solve

1 + 2 x + x = π + 2 πn : x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

1 + 2 x + x = π + 2 πn

Add similar elements:

2 x + x = 3 x

1 + 3 x = π + 2 πn

Move

1

to the right side

1 + 3 x = π + 2 πn

Subtract

1

from both sides

1 + 3 x - 1 = π + 2 πn - 1

Simplify

3 x = π + 2 πn - 1

3 x = π + 2 πn - 1

Divide both sides by

3

3 x = π + 2 πn - 1

Divide both sides by

3

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

Simplify

x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(-x)=tan(2x+1) ?
The general solution for tan(-x)=tan(2x+1) is x=(2pin)/3-1/3 ,x= pi/3+(2pin)/3-1/3