What is the general solution for y=(arcsin(1))/2 ?

The general solution for y=(arcsin(1))/2 is y= pi/4

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y=(arcsin(1))/2

y = \frac{arcsin ( 1 )}{2}

y = \frac{π}{4}

Decimal Notation

y = 0.78539 \dots

Solution steps

y = \frac{arcsin ( 1 )}{2}

Simplify

\frac{arcsin ( 1 )}{2} : \frac{π}{4}

\frac{arcsin ( 1 )}{2}

arcsin (1) = \frac{π}{2}

arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{2}

= \frac{\frac{π}{2}}{2}

Simplify

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

y = \frac{π}{4}

y = \frac{π}{4}

k, x^{2} + y^{2} = k \cdot arctan (\frac{y}{x})

t, cot (9 0^{0} + x) = tan (x)

t, arccos (\frac{( 5 cos ( x ) + 3 )}{( 5 + 3 cos ( x ) )}) = 2 arctan (\frac{1}{2} \cdot tan (\frac{x}{2}))

x, y = cos (\frac{y}{1.75}) + x - 0.5 y^{2}

c, 2 sin^{3} (x) + cos^{2} (x) - cos (x) = 0

What is the general solution for y=(arcsin(1))/2 ?
The general solution for y=(arcsin(1))/2 is y= pi/4