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

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

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(tan(x)+1)(cos(x)-1)=0

(tan (x) + 1) (cos (x) - 1) = 0

x = \frac{3 π}{4} + πn, x = 2 πn

Degrees

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

Solution steps

(tan (x) + 1) (cos (x) - 1) = 0

Solving each part separately

tan (x) + 1 = 0 or cos (x) - 1 = 0

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

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

cos (x) - 1 + 1 = 0 + 1

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = 2 πn

cot (θ) = - \frac{1}{3}

32 cos (θ) + 3 = 6

- 8 cos (θ) - 4 = - 4

2 cot (x) = 0

4 sinh (x^{3}) = 0

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