What is the general solution for 8tan(a)+3=2tan(a)+3 ?

The general solution for 8tan(a)+3=2tan(a)+3 is a=pin

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8tan(a)+3=2tan(a)+3

8 tan (a) + 3 = 2 tan (a) + 3

a = πn

Degrees

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

Solution steps

8 tan (a) + 3 = 2 tan (a) + 3

Solve by substitution

8 tan (a) + 3 = 2 tan (a) + 3

Let:

tan (a) = u

8 u + 3 = 2 u + 3

8 u + 3 = 2 u + 3 : u = 0

8 u + 3 = 2 u + 3

Subtract

3

from both sides

8 u + 3 - 3 = 2 u + 3 - 3

Simplify

8 u = 2 u

Move

2 u

to the left side

8 u = 2 u

Subtract

2 u

from both sides

8 u - 2 u = 2 u - 2 u

Simplify

6 u = 0

6 u = 0

Divide both sides by

6

6 u = 0

Divide both sides by

6

\frac{6 u}{6} = \frac{0}{6}

Simplify

u = 0

u = 0

Substitute back

u = tan (a)

tan (a) = 0

tan (a) = 0

tan (a) = 0 : a = πn

tan (a) = 0

General solutions for

tan (a) = 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}

a = 0 + πn

a = 0 + πn

Solve

a = 0 + πn : a = πn

a = 0 + πn

0 + πn = πn

a = πn

a = πn

Combine all the solutions

a = πn

tan (\frac{θ}{2}) = \frac{3}{3}

x, cos (x) + 1 = 0

- 8 sin (x) = 0

sin (θ + \frac{π}{4}) = 0

2 sin^{2} (x) - 7 sin (x) + 3 = 0, 0 \leq x \leq 2 π

What is the general solution for 8tan(a)+3=2tan(a)+3 ?
The general solution for 8tan(a)+3=2tan(a)+3 is a=pin