What is the general solution for -sin(c)+cos(c)+1=0 ?

The general solution for -sin(c)+cos(c)+1=0 is c= pi/2+2pin,c=pi+2pin

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-sin(c)+cos(c)+1=0

- sin (c) + cos (c) + 1 = 0

c = \frac{π}{2} + 2 πn, c = π + 2 πn

Solution steps

- sin (c) + cos (c) + 1 = 0

Rewrite using trig identities

1 - 2 sin (\frac{4 c - π}{4}) = 0

Move

1

to the right side

- 2 sin (\frac{4 c - π}{4}) = - 1

Divide both sides by

- 2

sin (\frac{4 c - π}{4}) = \frac{2}{2}

General solutions for

sin (\frac{4 c - π}{4}) = \frac{2}{2}

\frac{4 c - π}{4} = \frac{π}{4} + 2 πn, \frac{4 c - π}{4} = \frac{3 π}{4} + 2 πn

Solve

\frac{4 c - π}{4} = \frac{π}{4} + 2 πn : c = \frac{π}{2} + 2 πn

Solve

\frac{4 c - π}{4} = \frac{3 π}{4} + 2 πn : c = π + 2 πn

c = \frac{π}{2} + 2 πn, c = π + 2 πn

cot (a) = \frac{7}{51}

sin (x) = \frac{39}{44}

arctan (0.5) - arctan (\frac{1}{3}) = arctan (x)

arctan (x) = 3 0^{\circ}

x, - cos (x) + \frac{1}{y} = c

What is the general solution for -sin(c)+cos(c)+1=0 ?
The general solution for -sin(c)+cos(c)+1=0 is c= pi/2+2pin,c=pi+2pin