What is the general solution for sin(30)= x/1 ?

The general solution for sin(30)= x/1 is x= 1/2

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sin(30)= x/1

sin (3 0^{\circ}) = \frac{x}{1}

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

sin (3 0^{\circ}) = \frac{x}{1}

Switch sides

\frac{x}{1} = sin (3 0^{\circ})

Simplify

\frac{x}{1} : x

\frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

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}

= \frac{1}{2}

x = \frac{1}{2}

i, sin (x) + sin^{2} (x) + cos^{3} (x) = 1

x (15) = 0 + 705 \cdot 15 \cdot cos (2 5^{\circ})

t, 4 r + 12 s + 9 t = e^{3} x - 2 y + x y + cos (2 x + 3 y)

i, 5 sin (x) - 2 = 3 (1 - sin (x)) tan^{2} (x)

y, (b) y = e^{5 x} tan (x^{2} + 2)

What is the general solution for sin(30)= x/1 ?
The general solution for sin(30)= x/1 is x= 1/2