What is the general solution for tan(2θ)=(-1)/1 ?

The general solution for tan(2θ)=(-1)/1 is θ=(3pi)/8+(pin)/2

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tan(2θ)=(-1)/1

tan (2 θ) = \frac{- 1}{1}

θ = \frac{3 π}{8} + \frac{πn}{2}

Degrees

θ = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) = \frac{- 1}{1}

Apply rule

\frac{a}{1} = a

tan (2 θ) = - 1

General solutions for

tan (2 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}

2 θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{4} + πn

Solve

2 θ = \frac{3 π}{4} + πn : θ = \frac{3 π}{8} + \frac{πn}{2}

2 θ = \frac{3 π}{4} + πn

Divide both sides by

2

2 θ = \frac{3 π}{4} + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2} : \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

cos^{2} (\frac{x}{2}) = \frac{1}{2}

5^{5 - 3 c s c (θ)} = 1

sin (z) = \frac{4}{7}

2 cos (θ) - 2 cos (3 θ) = 0

5 - 7 sin (A) - 2 cos^{2} (A) = 0

What is the general solution for tan(2θ)=(-1)/1 ?
The general solution for tan(2θ)=(-1)/1 is θ=(3pi)/8+(pin)/2