What is the general solution for tan(2x+10)=cot(x-40) ?

The general solution for tan(2x+10)=cot(x-40) is x=40+(360n)/3

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tan(2x+10)=cot(x-40)

tan (2 x + 1 0^{\circ}) = cot (x - 4 0^{\circ})

x = 4 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solution steps

tan (2 x + 1 0^{\circ}) = cot (x - 4 0^{\circ})

Subtract

cot (x - 4 0^{\circ})

from both sides

tan (2 x + 1 0^{\circ}) - cot (x - 4 0^{\circ}) = 0

Simplify

tan (2 x + 1 0^{\circ}) - cot (x - 4 0^{\circ}) : tan (\frac{36 x + 18 0 ^{\circ}}{18}) - cot (\frac{9 x - 36 0 ^{\circ}}{9})

tan (\frac{36 x + 18 0 ^{\circ}}{18}) - cot (\frac{9 x - 36 0 ^{\circ}}{9}) = 0

Express with sin, cos

\frac{- cos ( \frac{- 36 0 ^{\circ} + 9 x}{9} ) cos ( \frac{18 0 ^{\circ} + 36 x}{18} ) + sin ( \frac{- 36 0 ^{\circ} + 9 x}{9} ) sin ( \frac{18 0 ^{\circ} + 36 x}{18} )}{cos ( \frac{18 0 ^{\circ} + 36 x}{18} ) sin ( \frac{- 36 0 ^{\circ} + 9 x}{9} )} = 0

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

- cos (\frac{- 36 0 ^{\circ} + 9 x}{9}) cos (\frac{18 0 ^{\circ} + 36 x}{18}) + sin (\frac{- 36 0 ^{\circ} + 9 x}{9}) sin (\frac{18 0 ^{\circ} + 36 x}{18}) = 0

Rewrite using trig identities

- cos (\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18}) = 0

Divide both sides by

- 1

cos (\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18}) = 0

General solutions for

cos (\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18}) = 0

\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18} = 9 0^{\circ} + 36 0^{\circ} n, \frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18} = 27 0^{\circ} + 36 0^{\circ} n

Solve

\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18} = 9 0^{\circ} + 36 0^{\circ} n : x = 4 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solve

\frac{- 36 0 ^{\circ} + 9 x}{9} + \frac{18 0 ^{\circ} + 36 x}{18} = 27 0^{\circ} + 36 0^{\circ} n : x = 10 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

x = 4 0^{\circ} + \frac{36 0 ^{\circ} n}{3}, x = 10 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Since the equation is undefined for:

10 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

x = 4 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

6 cos (2 t) = 0

2 cos^{2} (t) - 3 cos (t) = - 1, 0 \leq t \leq 2 π

4 cos (4 x) = 3

tan (y) = - 2

sin (θ) = - \frac{2}{2}

What is the general solution for tan(2x+10)=cot(x-40) ?
The general solution for tan(2x+10)=cot(x-40) is x=40+(360n)/3