What is the general solution for 40=tan(ln(18+1)+C) ?

The general solution for 40=tan(ln(18+1)+C) is C=1.54580…+pin-ln(19)

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40=tan(ln(18+1)+C)

40 = tan (ln (18 + 1) + C)

C = 1.54580 \dots + πn - ln (19)

Degrees

C = - 80.13602 \dots^{\circ} + 18 0^{\circ} n

Solution steps

40 = tan (ln (18 + 1) + C)

Switch sides

tan (ln (18 + 1) + C) = 40

Apply trig inverse properties

tan (ln (18 + 1) + C) = 40

General solutions for

tan (ln (18 + 1) + C) = 40

tan (x) = a \Rightarrow x = arctan (a) + πn

ln (18 + 1) + C = arctan (40) + πn

ln (18 + 1) + C = arctan (40) + πn

Solve

ln (18 + 1) + C = arctan (40) + πn : C = arctan (40) + πn - ln (19)

ln (18 + 1) + C = arctan (40) + πn

Add the numbers:

18 + 1 = 19

ln (19) + C = arctan (40) + πn

Move

ln (19)

to the right side

ln (19) + C = arctan (40) + πn

Subtract

ln (19)

from both sides

ln (19) + C - ln (19) = arctan (40) + πn - ln (19)

Simplify

C = arctan (40) + πn - ln (19)

C = arctan (40) + πn - ln (19)

C = arctan (40) + πn - ln (19)

Show solutions in decimal form

C = 1.54580 \dots + πn - ln (19)

tan (a) = - 26

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

5 sin (θ + 0.4636) = 1

0 = \frac{20}{3} sin (10 t) + 5 cos (10 t)

y, arctan (y) = \frac{t ^{2}}{2}

What is the general solution for 40=tan(ln(18+1)+C) ?
The general solution for 40=tan(ln(18+1)+C) is C=1.54580…+pin-ln(19)