What is the general solution for tan(5θ+30)=1 ?

The general solution for tan(5θ+30)=1 is θ=-6+(pin)/5+pi/(20)

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tan(5θ+30)=1

tan (5 θ + 30) = 1

θ = - 6 + \frac{πn}{5} + \frac{π}{20}

Degrees

θ = - 334.77467 \dots^{\circ} + 3 6^{\circ} n

Solution steps

tan (5 θ + 30) = 1

General solutions for

tan (5 θ + 30) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

5 θ + 30 = \frac{π}{4} + πn

5 θ + 30 = \frac{π}{4} + πn

Solve

5 θ + 30 = \frac{π}{4} + πn : θ = - 6 + \frac{πn}{5} + \frac{π}{20}

5 θ + 30 = \frac{π}{4} + πn

Move

30

to the right side

5 θ + 30 = \frac{π}{4} + πn

Subtract

30

from both sides

5 θ + 30 - 30 = \frac{π}{4} + πn - 30

Simplify

5 θ = \frac{π}{4} + πn - 30

5 θ = \frac{π}{4} + πn - 30

Divide both sides by

5

5 θ = \frac{π}{4} + πn - 30

Divide both sides by

5

\frac{5 θ}{5} = \frac{\frac{π}{4}}{5} + \frac{πn}{5} - \frac{30}{5}

Simplify

\frac{5 θ}{5} = \frac{\frac{π}{4}}{5} + \frac{πn}{5} - \frac{30}{5}

Simplify

\frac{5 θ}{5} : θ

\frac{5 θ}{5}

Divide the numbers:

\frac{5}{5} = 1

= θ

Simplify

\frac{\frac{π}{4}}{5} + \frac{πn}{5} - \frac{30}{5} : - 6 + \frac{πn}{5} + \frac{π}{20}

\frac{\frac{π}{4}}{5} + \frac{πn}{5} - \frac{30}{5}

Group like terms

= - \frac{30}{5} + \frac{πn}{5} + \frac{\frac{π}{4}}{5}

\frac{30}{5} = 6

\frac{30}{5}

Divide the numbers:

\frac{30}{5} = 6

= 6

\frac{\frac{π}{4}}{5} = \frac{π}{20}

\frac{\frac{π}{4}}{5}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 5}

Multiply the numbers:

4 \cdot 5 = 20

= \frac{π}{20}

= - 6 + \frac{πn}{5} + \frac{π}{20}

θ = - 6 + \frac{πn}{5} + \frac{π}{20}

θ = - 6 + \frac{πn}{5} + \frac{π}{20}

θ = - 6 + \frac{πn}{5} + \frac{π}{20}

θ = - 6 + \frac{πn}{5} + \frac{π}{20}

885 cos (θ) - 70 = cos (25 0^{\circ})

tan (x) + 5 = 0, x \leq 0 < 2 π

\frac{31.5}{sin ( 57.2 2 ^{\circ} )} = \frac{24.4}{sin ( x )}

23 cos (θ - \frac{π}{3}) = 3, 0 \leq θ \leq 2 π

sin (6 x) = 1

What is the general solution for tan(5θ+30)=1 ?
The general solution for tan(5θ+30)=1 is θ=-6+(pin)/5+pi/(20)