What is the general solution for arcsin(x)=arctan(1) ?

The general solution for arcsin(x)=arctan(1) is x=(sqrt(2))/2

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arcsin(x)=arctan(1)

arcsin (x) = arctan (1)

x = \frac{2}{2}

Solution steps

arcsin (x) = arctan (1)

arctan (1) = \frac{π}{4}

arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= \frac{π}{4}

arcsin (x) = \frac{π}{4}

Apply trig inverse properties

arcsin (x) = \frac{π}{4}

arcsin (x) = a \Rightarrow x = sin (a)

x = sin (\frac{π}{4})

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

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}

= \frac{2}{2}

= \frac{2}{2}

x = \frac{2}{2}

x = \frac{2}{2}

sec (θ) - 2 tan (θ) = 0

2 sin (θ) cos (θ) + 3 cos (θ) = 0

2 sin (x - \frac{π}{3}) = - 2

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

3 sin (θ) = 1 + cos (θ)

What is the general solution for arcsin(x)=arctan(1) ?
The general solution for arcsin(x)=arctan(1) is x=(sqrt(2))/2