What is the general solution for 5tan^2(x)-15=0 ?

The general solution for 5tan^2(x)-15=0 is x= pi/3+pin,x=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5tan^2(x)-15=0

5 tan^{2} (x) - 15 = 0

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

Degrees

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

Solution steps

5 tan^{2} (x) - 15 = 0

Solve by substitution

5 tan^{2} (x) - 15 = 0

Let:

tan (x) = u

5 u^{2} - 15 = 0

5 u^{2} - 15 = 0 : u = 3, u = - 3

5 u^{2} - 15 = 0

Move

15

to the right side

5 u^{2} - 15 = 0

Add

15

to both sides

5 u^{2} - 15 + 15 = 0 + 15

Simplify

5 u^{2} = 15

5 u^{2} = 15

Divide both sides by

5

5 u^{2} = 15

Divide both sides by

5

\frac{5 u ^{2}}{5} = \frac{15}{5}

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 (x)

tan (x) = 3, tan (x) = - 3

tan (x) = 3, tan (x) = - 3

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 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}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

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

tan (x) = - 3

General solutions for

tan (x) = - 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}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Combine all the solutions

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

- 2 cos^{2} (x) = - cos (x) - 1

sin (θ) = - \frac{2}{5}

- sin^{2} (x) = 2 cos (x) - 2

3 sin (2 θ) - 5 sin (θ) = 0

cos (2 x) = cos (x), 0 \leq x \leq 2 π

What is the general solution for 5tan^2(x)-15=0 ?
The general solution for 5tan^2(x)-15=0 is x= pi/3+pin,x=(2pi)/3+pin