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

The general solution for (tan(θ)+1)(sec(θ)-1)=0 is θ=(3pi)/4+pin,θ=2pin

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(tan(θ)+1)(sec(θ)-1)=0

(tan (θ) + 1) (sec (θ) - 1) = 0

θ = \frac{3 π}{4} + πn, θ = 2 πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

(tan (θ) + 1) (sec (θ) - 1) = 0

Solving each part separately

tan (θ) + 1 = 0 or sec (θ) - 1 = 0

tan (θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

tan (θ) + 1 = 0

Move

1

to the right side

tan (θ) + 1 = 0

Subtract

1

from 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

sec (θ) - 1 = 0 : θ = 2 πn

sec (θ) - 1 = 0

Move

1

to the right side

sec (θ) - 1 = 0

Add

1

to both sides

sec (θ) - 1 + 1 = 0 + 1

Simplify

sec (θ) = 1

sec (θ) = 1

General solutions for

sec (θ) = 1

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = \frac{3 π}{4} + πn, θ = 2 πn

sin (x) = - \frac{3}{5}

2 tan (x) cos (x) - tan (x) = 0

sin (\frac{x}{2}) = 2 - sin (\frac{x}{2})

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

sin (x) tan (x) = sin (x)

What is the general solution for (tan(θ)+1)(sec(θ)-1)=0 ?
The general solution for (tan(θ)+1)(sec(θ)-1)=0 is θ=(3pi)/4+pin,θ=2pin