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

The general solution for 1-cos(x)+sin(x)-1=0 is x= pi/4+pin

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

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

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{1 - cos ( x ) + sin ( x ) - 1}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

\frac{sin ( x )}{cos ( x )} - 1 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) - 1 = 0

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 1 = 0 + 1

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

3 tan (x^{3}) cot (x) 1 = 0

x, sin (\frac{y}{x}) + cos (\frac{x}{y}) = 0

sin (2 t) cos (t) = 0, 0 \leq t \leq π

2 sin^{3} (x) = sin^{3} (x)

θ, y = 4 sin (θ)

What is the general solution for 1-cos(x)+sin(x)-1=0 ?
The general solution for 1-cos(x)+sin(x)-1=0 is x= pi/4+pin