What is the general solution for 15sin(x)+19=14sin(x)+18 ?

The general solution for 15sin(x)+19=14sin(x)+18 is x=(3pi)/2+2pin

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15sin(x)+19=14sin(x)+18

15 sin (x) + 19 = 14 sin (x) + 18

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

15 sin (x) + 19 = 14 sin (x) + 18

Solve by substitution

15 sin (x) + 19 = 14 sin (x) + 18

Let:

sin (x) = u

15 u + 19 = 14 u + 18

15 u + 19 = 14 u + 18 : u = - 1

15 u + 19 = 14 u + 18

Move

19

to the right side

15 u + 19 = 14 u + 18

Subtract

19

from both sides

15 u + 19 - 19 = 14 u + 18 - 19

Simplify

15 u = 14 u - 1

15 u = 14 u - 1

Move

14 u

to the left side

15 u = 14 u - 1

Subtract

14 u

from both sides

15 u - 14 u = 14 u - 1 - 14 u

Simplify

u = - 1

u = - 1

Substitute back

u = sin (x)

sin (x) = - 1

sin (x) = - 1

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

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

3 + 3 sin (θ) = \frac{6}{3 - 2 sin ( θ )}

cos (x) = - 0.07

2 tan (θ) + 9 = 0

cos (x) = - 0.73

5 cos^{2} (x) - 3 cos (x) = 2

What is the general solution for 15sin(x)+19=14sin(x)+18 ?
The general solution for 15sin(x)+19=14sin(x)+18 is x=(3pi)/2+2pin