What is the general solution for 5+2sin(x)-7=0 ?

The general solution for 5+2sin(x)-7=0 is x= pi/2+2pin

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5+2sin(x)-7=0

5 + 2 sin (x) - 7 = 0

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

5 + 2 sin (x) - 7 = 0

Group like terms

2 sin (x) + 5 - 7 = 0

Add/Subtract the numbers:

5 - 7 = - 2

2 sin (x) - 2 = 0

Move

2

to the right side

2 sin (x) - 2 = 0

Add

2

to both sides

2 sin (x) - 2 + 2 = 0 + 2

Simplify

2 sin (x) = 2

2 sin (x) = 2

Divide both sides by

2

2 sin (x) = 2

Divide both sides by

2

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

0 = 4 sin (4 x + 3 π)

0.5 = 1 - sin (θ)

θ, r = 6 sin (2 θ)

1.5 sin (3 0^{\circ}) = sin (x)

3 sin (x) - 3 cos (x) = 6

What is the general solution for 5+2sin(x)-7=0 ?
The general solution for 5+2sin(x)-7=0 is x= pi/2+2pin