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= pi/4+pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{π}{4} + πn, x = π + 2 πn

Degrees

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

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to 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{π}{4} + πn

x = \frac{π}{4} + πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = \frac{π}{4} + πn, x = π + 2 πn

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

cos (2 x - \frac{π}{6}) = \frac{3}{2}

12 cos^{2} (θ) = 9

sin (\frac{x}{2}) - 1 = 0

sec (\frac{3 x}{2}) = - 2

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= pi/4+pin,x=pi+2pin