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

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

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solvefor θ,tan(2θ)=-1

solve for

θ, tan (2 θ) = - 1

θ = \frac{3 π}{8} + \frac{πn}{2}

Degrees

θ = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

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}

3 cos (z + 1.2) = 2

\frac{1}{2} = \frac{3}{2} \cdot cos (x)

cos (\frac{x}{2}) = - 12

4 sin (β) + 2 = 3 + 2 sin (β)

3 sin^{2} (x) = cos (x)

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