What is the general solution for (tan(θ)-1)(cos(θ)+1)=0 ?

The general solution for (tan(θ)-1)(cos(θ)+1)=0 is θ= pi/4+pin,θ=pi+2pin

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(tan(θ)-1)(cos(θ)+1)=0

(tan (θ) - 1) (cos (θ) + 1) = 0

θ = \frac{π}{4} + πn, θ = π + 2 πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

(tan (θ) - 1) (cos (θ) + 1) = 0

Solving each part separately

tan (θ) - 1 = 0 or cos (θ) + 1 = 0

tan (θ) - 1 = 0 : θ = \frac{π}{4} + πn

tan (θ) - 1 = 0

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

cos (θ) + 1 = 0 : θ = π + 2 πn

cos (θ) + 1 = 0

Move

1

to the right side

cos (θ) + 1 = 0

Subtract

1

from both sides

cos (θ) + 1 - 1 = 0 - 1

Simplify

cos (θ) = - 1

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = π + 2 πn

sin^{2} (θ) = cos^{2} (θ)

x, sin (x) = 0

cos (θ) = \frac{2}{5}

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

sin (y) = 0

What is the general solution for (tan(θ)-1)(cos(θ)+1)=0 ?
The general solution for (tan(θ)-1)(cos(θ)+1)=0 is θ= pi/4+pin,θ=pi+2pin