What is the general solution for tan(θ)-2cos(θ)tan(θ)=0 ?

The general solution for tan(θ)-2cos(θ)tan(θ)=0 is θ=pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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tan(θ)-2cos(θ)tan(θ)=0

tan (θ) - 2 cos (θ) tan (θ) = 0

θ = πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (θ) - 2 cos (θ) tan (θ) = 0

Factor

tan (θ) - 2 cos (θ) tan (θ) : - tan (θ) (2 cos (θ) - 1)

tan (θ) - 2 cos (θ) tan (θ)

Factor out common term

- tan (θ)

= - tan (θ) (- 1 + 2 cos (θ))

- tan (θ) (2 cos (θ) - 1) = 0

Solving each part separately

tan (θ) = 0 or 2 cos (θ) - 1 = 0

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

2 cos (θ) - 1 = 0 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 cos (θ) - 1 = 0

Move

1

to the right side

2 cos (θ) - 1 = 0

Add

1

to both sides

2 cos (θ) - 1 + 1 = 0 + 1

Simplify

2 cos (θ) = 1

2 cos (θ) = 1

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

tan (x) = sec^{2} (x) - 1

5 cos (x) = - 2 sin^{2} (x) + 4, 0 \leq x < 2 π

tan (x) = \frac{4 3}{4}

\frac{sin ^{2} ( x )}{1 - cos ( x )} = 1.23

x, 3 y = tan (6 x)

What is the general solution for tan(θ)-2cos(θ)tan(θ)=0 ?
The general solution for tan(θ)-2cos(θ)tan(θ)=0 is θ=pin,θ= pi/3+2pin,θ=(5pi)/3+2pin