What is the general solution for 2sin(θ)-2=0 ?

The general solution for 2sin(θ)-2=0 is θ= pi/2+2pin

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2sin(θ)-2=0

2 sin (θ) - 2 = 0

θ = \frac{π}{2} + 2 πn

Solution steps

2 sin (θ) - 2 = 0

Move

2

to the right side

2 sin (θ) = 2

Divide both sides by

2

sin (θ) = 1

General solutions for

sin (θ) = 1

θ = \frac{π}{2} + 2 πn

- 3 sin (θ) + 3 = 2 cos^{2} (θ)

2 sin (x - \frac{π}{2}) + 1 = 0

2 cos^{2} (x) - 1 = cos (x)

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

csc^{2} (x) + 6 csc (x) - 7 = 0

What is the general solution for 2sin(θ)-2=0 ?
The general solution for 2sin(θ)-2=0 is θ= pi/2+2pin