y=arccos(4x), so cosy=4x and -sinydy/dx=4.
At x=1/8, y=(pi)/3, therefore -sinydy/dx=4 and -sqrt(3)/2dy/dx=4, making dy/dx=-8/sqrt(3)=-8sqrt(3)/3.
a) The equation of the tangent is of the form y=mx+b where m=dy/dx=-8sqrt(3)/3. We know the tangent line must pass through (1/8,(pi)/3), so (pi)/3=-8sqrt(3)/3*1/8+b, so b=(pi)/3+sqrt(3)/3 and y=-8xsqrt(3)/3+(pi)/3+sqrt(3)/3 or 3y=sqrt(3)(1-8x)+(pi).
b) The equation of the normal has a gradient, or slope, of -1/m=sqrt(3)/8 and the equation of the normal is y=xsqrt(3)/8+c. Put the point (1/8,(pi)/3) in this equation: (pi)/3=sqrt(3)/64+c so c=(pi)/3-sqrt(3)/64 and y=xsqrt(3)/8-sqrt(3)/64+(pi)/3=sqrt(3)(8x-1)/64+(pi)/3.