What is the general solution for 5tan^2(A)-7=8 ?

The general solution for 5tan^2(A)-7=8 is A= pi/3+pin,A=(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(A)-7=8

5 tan^{2} (A) - 7 = 8

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

Degrees

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

Solution steps

5 tan^{2} (A) - 7 = 8

Solve by substitution

5 tan^{2} (A) - 7 = 8

Let:

tan (A) = u

5 u^{2} - 7 = 8

5 u^{2} - 7 = 8 : u = 3, u = - 3

5 u^{2} - 7 = 8

Move

7

to the right side

5 u^{2} - 7 = 8

Add

7

to both sides

5 u^{2} - 7 + 7 = 8 + 7

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 (A)

tan (A) = 3, tan (A) = - 3

tan (A) = 3, tan (A) = - 3

tan (A) = 3 : A = \frac{π}{3} + πn

tan (A) = 3

General solutions for

tan (A) = 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}

A = \frac{π}{3} + πn

A = \frac{π}{3} + πn

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

tan (A) = - 3

General solutions for

tan (A) = - 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}

A = \frac{2 π}{3} + πn

A = \frac{2 π}{3} + πn

Combine all the solutions

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

tan (x) = \frac{28}{16}

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

9 tan (2 x) - 18 cos (x) = 0

8 cos (2 t) - 14 sin (2 t) = 0

1 - 2 sin (θ) = 0

What is the general solution for 5tan^2(A)-7=8 ?
The general solution for 5tan^2(A)-7=8 is A= pi/3+pin,A=(2pi)/3+pin