What is the general solution for 8tan(B)+12=3tan(B)+7 ?

The general solution for 8tan(B)+12=3tan(B)+7 is B=(3pi)/4+pin

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8tan(B)+12=3tan(B)+7

8 tan (B) + 12 = 3 tan (B) + 7

B = \frac{3 π}{4} + πn

Degrees

B = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

8 tan (B) + 12 = 3 tan (B) + 7

Solve by substitution

8 tan (B) + 12 = 3 tan (B) + 7

Let:

tan (B) = u

8 u + 12 = 3 u + 7

8 u + 12 = 3 u + 7 : u = - 1

8 u + 12 = 3 u + 7

Move

12

to the right side

8 u + 12 = 3 u + 7

Subtract

12

from both sides

8 u + 12 - 12 = 3 u + 7 - 12

Simplify

8 u = 3 u - 5

8 u = 3 u - 5

Move

3 u

to the left side

8 u = 3 u - 5

Subtract

3 u

from both sides

8 u - 3 u = 3 u - 5 - 3 u

Simplify

5 u = - 5

5 u = - 5

Divide both sides by

5

5 u = - 5

Divide both sides by

5

\frac{5 u}{5} = \frac{- 5}{5}

Simplify

u = - 1

u = - 1

Substitute back

u = tan (B)

tan (B) = - 1

tan (B) = - 1

tan (B) = - 1 : B = \frac{3 π}{4} + πn

tan (B) = - 1

General solutions for

tan (B) = - 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}

B = \frac{3 π}{4} + πn

B = \frac{3 π}{4} + πn

Combine all the solutions

B = \frac{3 π}{4} + πn

- 2 + sin (2 θ) = - \frac{3}{2}

tan (c) = \frac{24}{7}

sin (t) = \frac{- 1}{2}

sin (x + \frac{π}{6}) + cos (x + \frac{π}{3}) = cos (2 x)

24 sin (2 t) - 24 cos (t) = 0

What is the general solution for 8tan(B)+12=3tan(B)+7 ?
The general solution for 8tan(B)+12=3tan(B)+7 is B=(3pi)/4+pin