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

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

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2+2cos(θ)=2-2sin(θ)

2 + 2 cos (θ) = 2 - 2 sin (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 + 2 cos (θ) = 2 - 2 sin (θ)

Subtract

2 - 2 sin (θ)

from both sides

2 cos (θ) + 2 sin (θ) = 0

Rewrite using trig identities

2 cos (θ) + 2 sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{2 cos ( θ ) + 2 sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

2 + \frac{2 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

2 + 2 tan (θ) = 0

2 + 2 tan (θ) = 0

Move

2

to the right side

2 + 2 tan (θ) = 0

Subtract

2

from both sides

2 + 2 tan (θ) - 2 = 0 - 2

Simplify

2 tan (θ) = - 2

2 tan (θ) = - 2

Divide both sides by

2

2 tan (θ) = - 2

Divide both sides by

2

\frac{2 tan ( θ )}{2} = \frac{- 2}{2}

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 (b) = 0.7245

cos (x) = \frac{6}{61}

sin (a + 3 0^{\circ}) = - \frac{3}{2}

cos (2 a) = 1

cos (\frac{2}{3} x) = 0

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