Question

Find the critical numbers sin^2x+cosx 0 < x< 2pi

Answer 1

`\bf sin^2(x)+cos(x)\iff [sin(x)]^2+cos(x)\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=2sin(x)cos(x)-sin(x)\implies 0=2sin(x)cos(x)-sin(x) \\\\\\ 0=sin(x)[2cos(x)-1]\implies \begin{cases} 0=sin(x)\\ sin^(-1)(0)=\measuredangle x\\ \pi \\ ----------\\ 0=2cos(x)-1\\\\ (1)/(2)=cos(x)\\\\ cos^(-1)\left( (1)/(2) \right)=\measuredangle x\\\\ (\pi )/(3),(5\pi )/(3) \end{cases}`