What is the general solution for sin(x)+2=3,0<x<360 ?

The general solution for sin(x)+2=3,0<x<360 is x=90

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sin(x)+2=3,0<x<360

sin (x) + 2 = 3, 0^{\circ} < x < 36 0^{\circ}

x = 9 0^{\circ}

Radians

x = \frac{π}{2}

Solution steps

sin (x) + 2 = 3, 0^{\circ} < x < 36 0^{\circ}

Move

2

to the right side

sin (x) + 2 = 3

Subtract

2

from both sides

sin (x) + 2 - 2 = 3 - 2

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 < x < 36 0^{\circ}

x = 9 0^{\circ}

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

sin (2 x) - 1 = cos (2 x)

- 2 sin (t) + 2 cos (2 t) = 0

sin (2 x) = sin (x), 0 \leq x \leq 2 π

sin (5 θ) = \frac{1}{2}

What is the general solution for sin(x)+2=3,0<x<360 ?
The general solution for sin(x)+2=3,0<x<360 is x=90