What is the general solution for 3-tan^2(b)=0 ?

The general solution for 3-tan^2(b)=0 is b= pi/3+pin,b=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3-tan^2(b)=0

3 - tan^{2} (b) = 0

b = \frac{π}{3} + πn, b = \frac{2 π}{3} + πn

Degrees

b = 6 0^{\circ} + 18 0^{\circ} n, b = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

3 - tan^{2} (b) = 0

Solve by substitution

3 - tan^{2} (b) = 0

Let:

tan (b) = u

3 - u^{2} = 0

3 - u^{2} = 0 : u = 3, u = - 3

3 - u^{2} = 0

Move

3

to the right side

3 - u^{2} = 0

Subtract

3

from both sides

3 - u^{2} - 3 = 0 - 3

Simplify

- u^{2} = - 3

- u^{2} = - 3

Divide both sides by

- 1

- u^{2} = - 3

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{- 3}{- 1}

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3, u = - 3

Substitute back

u = tan (b)

tan (b) = 3, tan (b) = - 3

tan (b) = 3, tan (b) = - 3

tan (b) = 3 : b = \frac{π}{3} + πn

tan (b) = 3

General solutions for

tan (b) = 3

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}

b = \frac{π}{3} + πn

b = \frac{π}{3} + πn

tan (b) = - 3 : b = \frac{2 π}{3} + πn

tan (b) = - 3

General solutions for

tan (b) = - 3

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}

b = \frac{2 π}{3} + πn

b = \frac{2 π}{3} + πn

Combine all the solutions

b = \frac{π}{3} + πn, b = \frac{2 π}{3} + πn

\frac{( 2 sin ( x ) ) ( sin ( x ) )}{sin ( x )} + 1 = 3

tan (2 θ) = - \frac{2}{5}

cos (2 x) - 3 cos (x) = - 2

3 cos (x) - 3 sin (x) = 0

cos^{2} (3 x) = cos (9 x^{2})

What is the general solution for 3-tan^2(b)=0 ?
The general solution for 3-tan^2(b)=0 is b= pi/3+pin,b=(2pi)/3+pin