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

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

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

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

x = π + 2 πn, x = \frac{π}{4} + πn

Degrees

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

Solution steps

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

Solving each part separately

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

1 + cos (x) = 0 : x = π + 2 πn

1 + cos (x) = 0

Move

1

to the right side

1 + cos (x) = 0

Subtract

1

from both sides

1 + cos (x) - 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 = π + 2 πn

x = π + 2 πn

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

tan (x) - 1 = 0