What is the general solution for (5tan(θ)+25)=0 ?

The general solution for (5tan(θ)+25)=0 is θ=-1.37340…+pin

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(5tan(θ)+25)=0

(5 tan (θ) + 25) = 0

θ = - 1.37340 \dots + πn

Degrees

θ = - 78.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

(5 tan (θ) + 25) = 0

Move

25

to the right side

5 tan (θ) + 25 = 0

Subtract

25

from both sides

5 tan (θ) + 25 - 25 = 0 - 25

Simplify

5 tan (θ) = - 25

5 tan (θ) = - 25

Divide both sides by

5

5 tan (θ) = - 25

Divide both sides by

5

\frac{5 tan ( θ )}{5} = \frac{- 25}{5}

Simplify

\frac{5 tan ( θ )}{5} = \frac{- 25}{5}

Simplify

\frac{5 tan ( θ )}{5} : tan (θ)

\frac{5 tan ( θ )}{5}

Divide the numbers:

\frac{5}{5} = 1

= tan (θ)

Simplify

\frac{- 25}{5} : - 5

\frac{- 25}{5}

Apply the fraction rule:

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

= - \frac{25}{5}

Divide the numbers:

\frac{25}{5} = 5

= - 5

tan (θ) = - 5

tan (θ) = - 5

tan (θ) = - 5

Apply trig inverse properties

tan (θ) = - 5

General solutions for

tan (θ) = - 5

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 5) + πn

θ = arctan (- 5) + πn

Show solutions in decimal form

θ = - 1.37340 \dots + πn

4 = cos (x)

x, e^{4 y} = cos (4 x + y)

3 tan (2 x) - 3 cot (x) = 0

cos (θ) = (\frac{1}{4})

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

What is the general solution for (5tan(θ)+25)=0 ?
The general solution for (5tan(θ)+25)=0 is θ=-1.37340…+pin