What is the general solution for 12tan(B)+5=5tan(B)+5 ?

The general solution for 12tan(B)+5=5tan(B)+5 is B=pin

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12tan(B)+5=5tan(B)+5

12 tan (B) + 5 = 5 tan (B) + 5

B = πn

Degrees

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

Solution steps

12 tan (B) + 5 = 5 tan (B) + 5

Solve by substitution

12 tan (B) + 5 = 5 tan (B) + 5

Let:

tan (B) = u

12 u + 5 = 5 u + 5

12 u + 5 = 5 u + 5 : u = 0

12 u + 5 = 5 u + 5

Subtract

5

from both sides

12 u + 5 - 5 = 5 u + 5 - 5

Simplify

12 u = 5 u

Move

5 u

to the left side

12 u = 5 u

Subtract

5 u

from both sides

12 u - 5 u = 5 u - 5 u

Simplify

7 u = 0

7 u = 0

Divide both sides by

7

7 u = 0

Divide both sides by

7

\frac{7 u}{7} = \frac{0}{7}

Simplify

u = 0

u = 0

Substitute back

u = tan (B)

tan (B) = 0

tan (B) = 0

tan (B) = 0 : B = πn

tan (B) = 0

General solutions for

tan (B) = 0

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 = 0 + πn

B = 0 + πn

Solve

B = 0 + πn : B = πn

B = 0 + πn

0 + πn = πn

B = πn

B = πn

Combine all the solutions

B = πn

3 cos (A) = sin^{2} (A)

\frac{sin ( x )}{tan ( x )} = - 0.375

sin (2 x) = - sin (x), 0 \leq x \leq 2 π

2 sin (x) cos (x) + 2 sin (x) + cos (x) + 1 = 0

sin^{2} (θ) = 0

What is the general solution for 12tan(B)+5=5tan(B)+5 ?
The general solution for 12tan(B)+5=5tan(B)+5 is B=pin