What is the general solution for -1=tan(2(0)^3+C) ?

The general solution for -1=tan(2(0)^3+C) is C=(3pi)/4+pin

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-1=tan(2(0)^3+C)

- 1 = tan (2 (0)^{3} + C)

C = \frac{3 π}{4} + πn

Degrees

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

Solution steps

- 1 = tan (2 (0)^{3} + C)

Switch sides

tan (2 \cdot 0^{3} + C) = - 1

General solutions for

tan (2 0^{3} + C) = - 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}

2 0^{3} + C = \frac{3 π}{4} + πn

2 0^{3} + C = \frac{3 π}{4} + πn

Solve

2 0^{3} + C = \frac{3 π}{4} + πn : C = \frac{3 π}{4} + πn

2 \cdot 0^{3} + C = \frac{3 π}{4} + πn

Apply rule

0^{a} = 0

0^{3} = 0

2 \cdot 0 + C = \frac{3 π}{4} + πn

Apply rule

0 \cdot a = 0

0 + C = \frac{3 π}{4} + πn

0 + C = C

C = \frac{3 π}{4} + πn

C = \frac{3 π}{4} + πn

sin (x) = \frac{1}{2}, 0 < x < \frac{π}{2}

sin (2 x) = cos (8 x)

8 = sin (x)

arctan (4 x) + arctan (6 x) = - \frac{π}{2}

- 2 cot (x) + csc^{2} (x) = 0

What is the general solution for -1=tan(2(0)^3+C) ?
The general solution for -1=tan(2(0)^3+C) is C=(3pi)/4+pin