What is the general solution for 3+3cos(θ)=3+3sin(θ) ?

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

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3+3cos(θ)=3+3sin(θ)

3 + 3 cos (θ) = 3 + 3 sin (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

3 + 3 cos (θ) = 3 + 3 sin (θ)

Subtract

3 + 3 sin (θ)

from both sides

3 cos (θ) - 3 sin (θ) = 0

Rewrite using trig identities

3 cos (θ) - 3 sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{3 cos ( θ ) - 3 sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

3 - \frac{3 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

3 - 3 tan (θ) = 0

3 - 3 tan (θ) = 0

Move

3

to the right side

3 - 3 tan (θ) = 0

Subtract

3

from both sides

3 - 3 tan (θ) - 3 = 0 - 3

Simplify

- 3 tan (θ) = - 3

- 3 tan (θ) = - 3

Divide both sides by

- 3

- 3 tan (θ) = - 3

Divide both sides by

- 3

\frac{- 3 tan ( θ )}{- 3} = \frac{- 3}{- 3}

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{π}{4} + πn

θ = \frac{π}{4} + πn

sinh (x) = \frac{4}{3}

x, f^{- 1} = arctan (x^{3} - 1)

3 cos (B) + 3 = 0

sin (2 θ) = \frac{3}{4}

sin^{2} (x) = 2 \cdot sin (x) + 3

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