What is the general solution for tan(2θ)=1,0<= θ<= 360 ?

The general solution for tan(2θ)=1,0<= θ<= 360 is θ=22.5,θ=112.5,θ=202.5,θ=292.5

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tan(2θ)=1,0<= θ<= 360

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 22. 5^{\circ}, θ = 112. 5^{\circ}, θ = 202. 5^{\circ}, θ = 292. 5^{\circ}

Radians

θ = \frac{π}{8}, θ = \frac{5 π}{8}, θ = \frac{9 π}{8}, θ = \frac{13 π}{8}

Solution steps

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

General solutions for

tan (2 θ) = 1

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 4 5^{\circ} + 18 0^{\circ} n

Solve

2 θ = 4 5^{\circ} + 18 0^{\circ} n : θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 θ = 4 5^{\circ} + 18 0^{\circ} n

Divide both sides by

2

2 θ = 4 5^{\circ} + 18 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} : 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

\frac{4 5 ^{\circ}}{2} = 22. 5^{\circ}

\frac{4 5 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= 22. 5^{\circ}

= 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 22. 5^{\circ}, θ = 112. 5^{\circ}, θ = 202. 5^{\circ}, θ = 292. 5^{\circ}

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

3 tan (\frac{θ}{2}) - 1 = 0

tan (\frac{x}{2}) = 3, 0 \leq x < 2 π

sin (x) = \frac{4}{cos ( x )}

tan^{3} (x) - 2 tan (x) = tan (x)

What is the general solution for tan(2θ)=1,0<= θ<= 360 ?
The general solution for tan(2θ)=1,0<= θ<= 360 is θ=22.5,θ=112.5,θ=202.5,θ=292.5