What is the general solution for 1=tan(pi/4+c) ?

The general solution for 1=tan(pi/4+c) is c=pin

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1=tan(pi/4+c)

1 = tan (\frac{π}{4} + c)

c = πn

Degrees

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

Solution steps

1 = tan (\frac{π}{4} + c)

Switch sides

tan (\frac{π}{4} + c) = 1

General solutions for

tan (\frac{π}{4} + 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}

\frac{π}{4} + c = \frac{π}{4} + πn

\frac{π}{4} + c = \frac{π}{4} + πn

Solve

\frac{π}{4} + c = \frac{π}{4} + πn : c = πn

\frac{π}{4} + c = \frac{π}{4} + πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} + c - \frac{π}{4} = \frac{π}{4} + πn - \frac{π}{4}

Simplify

c = πn

c = πn

cos (θ) - cos (3 θ) = 0

- tan (2 θ) + 2 sin (θ) = 0, 0 \leq θ < 2 π

cos (x) = \frac{2 tan ( x )}{1 + tan ^{2} ( x )}

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

cos (2 x) - cos (6 x) - sin (4 x) = 0

What is the general solution for 1=tan(pi/4+c) ?
The general solution for 1=tan(pi/4+c) is c=pin