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larisa86 [58]
3 years ago
11

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

Mathematics
1 answer:
ryzh [129]3 years ago
5 0
\bf sin^2(x)+cos(x)\iff [sin(x)]^2+cos(x)\\\\&#10;-----------------------------\\\\&#10;\cfrac{dy}{dx}=2sin(x)cos(x)-sin(x)\implies 0=2sin(x)cos(x)-sin(x)&#10;\\\\\\&#10;0=sin(x)[2cos(x)-1]\implies &#10;\begin{cases}&#10;0=sin(x)\\&#10;sin^{-1}(0)=\measuredangle x\\&#10;\pi \\&#10;----------\\&#10;0=2cos(x)-1\\\\ \frac{1}{2}=cos(x)\\\\ cos^{-1}\left( \frac{1}{2} \right)=\measuredangle x\\\\ \frac{\pi }{3},\frac{5\pi }{3}&#10;\end{cases}
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