What is the general solution for 3tan(B)-4=tan(B)-2 ?

The general solution for 3tan(B)-4=tan(B)-2 is B= pi/4+pin

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3tan(B)-4=tan(B)-2

3 tan (B) - 4 = tan (B) - 2

B = \frac{π}{4} + πn

Degrees

B = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

3 tan (B) - 4 = tan (B) - 2

Solve by substitution

3 tan (B) - 4 = tan (B) - 2

Let:

tan (B) = u

3 u - 4 = u - 2

3 u - 4 = u - 2 : u = 1

3 u - 4 = u - 2

Move

4

to the right side

3 u - 4 = u - 2

Add

4

to both sides

3 u - 4 + 4 = u - 2 + 4

Simplify

3 u = u + 2

3 u = u + 2

Move

u

to the left side

3 u = u + 2

Subtract

u

from both sides

3 u - u = u + 2 - u

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

\frac{2 u}{2} = \frac{2}{2}

Simplify

u = 1

u = 1

Substitute back

u = tan (B)

tan (B) = 1

tan (B) = 1

tan (B) = 1 : B = \frac{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{π}{4} + πn

B = \frac{π}{4} + πn

Combine all the solutions

B = \frac{π}{4} + πn

sin (2 x) = 5 cos (2 x)

5 sec (2 x) + 2 = 0

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

1 sin (4 5^{\circ}) = 1.33 sin (θ)

3 cot^{2} (x) = 9

What is the general solution for 3tan(B)-4=tan(B)-2 ?
The general solution for 3tan(B)-4=tan(B)-2 is B= pi/4+pin