Answer
(a) $y=\pi$
(b) $x=0$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to both sides of the equation: $LHS=(0)^2cos^2\pi-sin(\pi)=0=RHS$
(a) Given the equation $x^2cos^2(y)-sin(y)=0$, differentiate both sides with respect to $x$: $2x\cdot cos^2(y)+2x^2cos(y)sin(y)y'-cos(y)y'=0$ and $y'=\frac{2x\cdot cos^2(y)}{cos(y)-x^2sin(2y)}$
Thus the slope of the tangent line at the given point is $m=y'=\frac{2(0)\cdot cos^2(\pi)}{cos(\pi)-0^2sin(2\pi)}=0$ and the equation for the tangent line is $y=\pi$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\infty$, and the line equation is $x=0$
