I am trying to find absolute maximum and absolute minimum values for$\ f(x)=\cos(x)+\sin(2x)$ on the interval $\ [0,\fracπ2]$.

I started off by finding the derivative:

$\ f'(x)=-\sin x+2\cos(2x)$

Then I tried to find the critical numbers. But that's where I ran amuck. I came up with:

$\sin x=2\cos(2x)$

How do I go about solving for x?


use that $$\cos(2x)=1-2\sin^2 (x)$$ plugging this in your equation we get $$-\sin(x)+2(1-2\sin^2 (x))=0$$ this is a quadratic equation in $\sin(x)$

  • $\begingroup$ Sorry, not sure I've understood how that helps. $\endgroup$ – rotaredom Nov 17 '17 at 19:13
  • $\begingroup$ does it help now? $\endgroup$ – Dr. Sonnhard Graubner Nov 17 '17 at 19:18
  • $\begingroup$ Ahh... I think I'm catching the idea now... $\endgroup$ – rotaredom Nov 17 '17 at 19:19
  • $\begingroup$ then good solving now $\endgroup$ – Dr. Sonnhard Graubner Nov 17 '17 at 19:19
  • $\begingroup$ Thanks for the help, got it! $\endgroup$ – rotaredom Nov 17 '17 at 20:10

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