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)?

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)?

espex · Answer

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

anonymous · Answer

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.

anonymous · Answer

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