What is the general solution for 1=tan(0+c) ?

The general solution for 1=tan(0+c) is c= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1=tan(0+c)

1 = tan (0 + c)

c = \frac{π}{4} + πn

Degrees

c = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

1 = tan (0 + c)

Switch sides

tan (0 + c) = 1

General solutions for

tan (0 + c) = 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}

0 + c = \frac{π}{4} + πn

0 + c = \frac{π}{4} + πn

Solve

0 + c = \frac{π}{4} + πn : c = \frac{π}{4} + πn

0 + c = \frac{π}{4} + πn

0 + c = c

c = \frac{π}{4} + πn

c = \frac{π}{4} + πn

4 sin^{2} (x) = 3 - 4 sin (x)

tan (θ) = - \frac{2}{2}

\frac{sin ( 7 8 ^{\circ} )}{14} = \frac{sin ( a )}{6}

tan (x) = (- \frac{6}{5})

cos (x^{2}) = 1

What is the general solution for 1=tan(0+c) ?
The general solution for 1=tan(0+c) is c= pi/4+pin