What is the general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c ?

The general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c is y=tan(c-1/2+t-1/2 t^2)

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solvefor y,arctan(y)+1/2 (1-t)^2=c

solve for

y, arctan (y) + \frac{1}{2} (1 - t)^{2} = c

y = tan (c - \frac{1}{2} + t - \frac{1}{2} t^{2})

Solution steps

arctan (y) + \frac{1}{2} (1 - t)^{2} = c

Move

\frac{1}{2} (1 - t)^{2}

to the right side

arctan (y) = c - \frac{1}{2} + t - \frac{1}{2} t^{2}

Apply trig inverse properties

y = tan (c - \frac{1}{2} + t - \frac{1}{2} t^{2})

\frac{15 - 3}{3} = 5 - \frac{2}{3} \cdot csc (4 x + \frac{π}{2})

tan (x) + 4 sin (x) = 0

1 + 2 \cdot sin (x) = 2

1 + 2 \cdot sin (x) = 3

sin (α) = \frac{120 sin ( 6 0 ^{\circ} )}{1200}

What is the general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c ?
The general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c is y=tan(c-1/2+t-1/2 t^2)