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

sin (θ) + 2 = - sin (θ)

cos (x) = π

3 cot (θ) + 1 = 0

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

cos^{4} (2 x) - sin^{4} (2 x) = 1

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