Question

2cosx+sin^2x finding critical values Finding the derivative and simplifying I have sinx=cosx, why is π a critical value. Isnt π/4(n)?

Answer 1

Given your equation, sin(x)=cos(x), pi is a critical value because the equation is false at pi.
sin(pi) != cos(pi), 0 != 1

Answer 2

I get for the derivitive: \[-2\sin x +2 \sin x \cos x\] simplifying to \[2\sin x(\cos x-1)\] which gives critical pts at 0 and \[pi\] when either sine or cosine are 0.

Answer 3

The red curve is the orginal equation; blue 2cos(x); brown \[\sin^{2} x\]