What is the general solution for sqrt(2)-cos(θ)=sqrt(2)+sin(θ) ?

The general solution for sqrt(2)-cos(θ)=sqrt(2)+sin(θ) is θ=(3pi)/4+pin

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sqrt(2)-cos(θ)=sqrt(2)+sin(θ)

2 - cos (θ) = 2 + sin (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 - cos (θ) = 2 + sin (θ)

Subtract

2 + sin (θ)

from both sides

- cos (θ) - sin (θ) = 0

Rewrite using trig identities

- cos (θ) - sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{- cos ( θ ) - sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

- 1 - \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

- 1 - tan (θ) = 0

- 1 - tan (θ) = 0

Move

1

to the right side

- 1 - tan (θ) = 0

Add

1

to both sides

- 1 - tan (θ) + 1 = 0 + 1

Simplify

- tan (θ) = 1

- tan (θ) = 1

Divide both sides by

- 1

- tan (θ) = 1

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{1}{- 1}

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

sin (x) = 6757

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

- 5 = - 5 + 4 sin (- 2 x + \frac{3 π}{4})

6 sec^{2} (x) + tan (x) - 9 = 0

3 tan^{2} (x) - cos^{2} (x) = sin^{2} (x)

What is the general solution for sqrt(2)-cos(θ)=sqrt(2)+sin(θ) ?
The general solution for sqrt(2)-cos(θ)=sqrt(2)+sin(θ) is θ=(3pi)/4+pin