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

The general solution for 1-sin(x)=1+cos(x) is x=(3pi)/4+pin

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

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

x = \frac{3 π}{4} + πn

Degrees

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

Solution steps

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

Subtract

1 + cos (x)

from both sides

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

Rewrite using trig identities

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

Divide both sides by

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

\frac{- sin ( x ) - cos ( x )}{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

Divide both sides by

- 1

- tan (x) = 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{1}{- 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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

3 cos (2 x) - 7 cos (x) + 5 = 0

6 cos (x) + 3 = 0

6 arccos (3 x) = 5 π

2 sin (θ) = - 3, 36 0^{\circ}, 18 0^{\circ} \leq θ

tan (θ) = \frac{- 3}{4}

What is the general solution for 1-sin(x)=1+cos(x) ?
The general solution for 1-sin(x)=1+cos(x) is x=(3pi)/4+pin