What is the general solution for 5tan(b)+7=2tan(b)+4 ?

The general solution for 5tan(b)+7=2tan(b)+4 is b=(3pi)/4+pin

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5tan(b)+7=2tan(b)+4

5 tan (b) + 7 = 2 tan (b) + 4

b = \frac{3 π}{4} + πn

Degrees

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

Solution steps

5 tan (b) + 7 = 2 tan (b) + 4

Solve by substitution

5 tan (b) + 7 = 2 tan (b) + 4

Let:

tan (b) = u

5 u + 7 = 2 u + 4

5 u + 7 = 2 u + 4 : u = - 1

5 u + 7 = 2 u + 4

Move

7

to the right side

5 u + 7 = 2 u + 4

Subtract

7

from both sides

5 u + 7 - 7 = 2 u + 4 - 7

Simplify

5 u = 2 u - 3

5 u = 2 u - 3

Move

2 u

to the left side

5 u = 2 u - 3

Subtract

2 u

from both sides

5 u - 2 u = 2 u - 3 - 2 u

Simplify

3 u = - 3

3 u = - 3

Divide both sides by

3

3 u = - 3

Divide both sides by

3

\frac{3 u}{3} = \frac{- 3}{3}

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

cot (x) = 1, 0 \leq x < 2 π

sin (4 x) cos (6 x) - cos (4 x) sin (6 x) = - 0.75

cos (2 x) = 0.685

4 tan^{2} (x) = 12

cos (θ) = - 0.71

What is the general solution for 5tan(b)+7=2tan(b)+4 ?
The general solution for 5tan(b)+7=2tan(b)+4 is b=(3pi)/4+pin