What is the general solution for 3+2sin(θ)=3+2cos(θ) ?

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

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3+2sin(θ)=3+2cos(θ)

3 + 2 sin (θ) = 3 + 2 cos (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

3 + 2 sin (θ) = 3 + 2 cos (θ)

Subtract

3 + 2 cos (θ)

from both sides

2 sin (θ) - 2 cos (θ) = 0

Rewrite using trig identities

2 sin (θ) - 2 cos (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

\frac{2 sin ( θ )}{cos ( θ )} - 2 = 0

Use the basic trigonometric identity:

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

2 tan (θ) - 2 = 0

2 tan (θ) - 2 = 0

Move

2

to the right side

2 tan (θ) - 2 = 0

Add

2

to both sides

2 tan (θ) - 2 + 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{π}{4} + πn

θ = \frac{π}{4} + πn

\frac{sin ( x )}{cos ( x )} = \frac{1}{3}

5 sin (θ) + 4 = 0

sin (\frac{5 x}{2}) = 0

sin (3 x - \frac{2 π}{3}) = - \frac{1}{2}

1 - cos (θ) = 1

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