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)

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

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

Solution steps

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

Rewrite using trig identities

- 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

\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

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

Divide both sides by

- 1

sin (1 + 2 x + x) = 0

General solutions for

sin (1 + 2 x + x) = 0

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}

Solve

1 + 2 x + x = π + 2 πn : 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}

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

sin (6 x) cos (9 x) - cos (6 x) sin (9 x) = - 0.4

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

cos^{3} (x) = cos (2 x + x)

x, (d^{2} - 3 d + 2) y = sin (e^{x})

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