What is the general solution for tan(x-20)=tan(2x+10) ?

The general solution for tan(x-20)=tan(2x+10) is x=-360n-20-10,x=-10-200-360n

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tan(x-20)=tan(2x+10)

tan (x - 2 0^{\circ}) = tan (2 x + 10)

x = - 36 0^{\circ} n - 2 0^{\circ} - 10, x = - 10 - 20 0^{\circ} - 36 0^{\circ} n

Solution steps

tan (x - 2 0^{\circ}) = tan (2 x + 10)

Subtract

tan (2 x + 10)

from both sides

tan (x - 2 0^{\circ}) - tan (2 x + 10) = 0

Express with sin, cos

\frac{cos ( 10 + 2 x ) sin ( \frac{- 18 0 ^{\circ} + 9 x}{9} ) - cos ( \frac{- 18 0 ^{\circ} + 9 x}{9} ) sin ( 10 + 2 x )}{cos ( 10 + 2 x ) cos ( \frac{- 18 0 ^{\circ} + 9 x}{9} )} = 0

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

cos (10 + 2 x) sin (\frac{- 18 0 ^{\circ} + 9 x}{9}) - cos (\frac{- 18 0 ^{\circ} + 9 x}{9}) sin (10 + 2 x) = 0

Rewrite using trig identities

sin (\frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x)) = 0

General solutions for

sin (\frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x)) = 0

\frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x) = 0 + 36 0^{\circ} n, \frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x) = 18 0^{\circ} + 36 0^{\circ} n

Solve

\frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x) = 0 + 36 0^{\circ} n : x = - 36 0^{\circ} n - 2 0^{\circ} - 10

Solve

\frac{- 18 0 ^{\circ} + 9 x}{9} - (10 + 2 x) = 18 0^{\circ} + 36 0^{\circ} n : x = - 10 - 20 0^{\circ} - 36 0^{\circ} n

x = - 36 0^{\circ} n - 2 0^{\circ} - 10, x = - 10 - 20 0^{\circ} - 36 0^{\circ} n

sin (5 x) cos (x) - cos (5 x) sin (x) = 0

2 sin (3 x + 1) + 1 = 2

2 sin (3 x + 1) + 1 = 1

sin (x) = 6946

sin (2 x) - \frac{3}{2} = 0, - 2 π \leq x \leq 2 π

What is the general solution for tan(x-20)=tan(2x+10) ?
The general solution for tan(x-20)=tan(2x+10) is x=-360n-20-10,x=-10-200-360n