What is the general solution for tan(θ)=sqrt(3)tan(45) ?

The general solution for tan(θ)=sqrt(3)tan(45) is θ=60+180n

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tan(θ)=sqrt(3)tan(45)

tan (θ) = 3 tan (4 5^{\circ})

θ = 6 0^{\circ} + 18 0^{\circ} n

Radians

θ = \frac{π}{3} + πn

Solution steps

tan (θ) = 3 tan (4 5^{\circ})

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

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}

= 1

= 1

tan (θ) = 3 \cdot 1

Multiply:

3 \cdot 1 = 3

tan (θ) = 3

General solutions for

tan (θ) = 3

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}

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

- cot (x) csc (x) = 0

2 cos (2 θ) - 3 cos (θ) - 5 = - 7

cos (x) tan (x) = - \frac{1}{2}

cos (3 x + π) = \frac{2}{2}

tan (θ) = - 16

What is the general solution for tan(θ)=sqrt(3)tan(45) ?
The general solution for tan(θ)=sqrt(3)tan(45) is θ=60+180n