What is the general solution for tan(3θ)=-1 ?

The general solution for tan(3θ)=-1 is θ= pi/4+(pin)/3

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tan(3θ)=-1

tan (3 θ) = - 1

θ = \frac{π}{4} + \frac{πn}{3}

Degrees

θ = 4 5^{\circ} + 6 0^{\circ} n

Solution steps

tan (3 θ) = - 1

General solutions for

tan (3 θ) = - 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}

3 θ = \frac{3 π}{4} + πn

3 θ = \frac{3 π}{4} + πn

Solve

3 θ = \frac{3 π}{4} + πn : θ = \frac{π}{4} + \frac{πn}{3}

3 θ = \frac{3 π}{4} + πn

Divide both sides by

3

3 θ = \frac{3 π}{4} + πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3} : \frac{π}{4} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} = \frac{π}{4}

\frac{\frac{3 π}{4}}{3}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{3 π}{12}

Cancel the common factor:

3

= \frac{π}{4}

= \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

1 + 2 sin (2 x) = 0

2 sech^{2} (x) + tanh (x) = 0

8 cos (2 θ) = - 24 cos (θ) - 16

16 cos^{2} (θ) - 9 = 0

sin (a) + cos (a) = 1

What is the general solution for tan(3θ)=-1 ?
The general solution for tan(3θ)=-1 is θ= pi/4+(pin)/3