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

The general solution for tan(2θ)=1,0<= θ<= 2pi is θ= pi/8 ,θ=(5pi)/8 ,θ=(9pi)/8 ,θ=(13pi)/8

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tan(2θ)=1,0<= θ<= 2pi

tan (2 θ) = 1, 0 \leq θ \leq 2 π

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

Degrees

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

Solution steps

tan (2 θ) = 1, 0 \leq θ \leq 2 π

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{π}{4} + πn

2 θ = \frac{π}{4} + πn

Solve

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

2 θ = \frac{π}{4} + πn

Divide both sides by

2

2 θ = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

Solutions for the range

0 \leq θ \leq 2 π

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

0.4 = 0.4 cos^{2} (θ)

sin (α) = \frac{15}{17}

4 sin^{2} (x) cot (x) - cot (x) = 0, 0 \leq x \leq 2 π

14 - sin (θ) = cos (2 θ)

csc (θ) = - \frac{5}{13}

What is the general solution for tan(2θ)=1,0<= θ<= 2pi ?
The general solution for tan(2θ)=1,0<= θ<= 2pi is θ= pi/8 ,θ=(5pi)/8 ,θ=(9pi)/8 ,θ=(13pi)/8