- #1

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## Homework Statement

I'm asked to find the max value of [itex]\sqrt{3}cosx+sinx[/itex]

## Homework Equations

[itex]y=sinx,y'=cosx[/itex]

[itex]y=cosx, y'=-sinx[/itex]

[itex]cos^2x=1-sin^2x[/itex]

## The Attempt at a Solution

I let [itex]y=\sqrt{3}cosx+sinx[/itex]

Hence, [tex]\frac{dy}{dx}=-\sqrt{3}sinx+cosx=0[/tex]

Attempting to solve this for x: [tex]\sqrt{3}sinx=cosx \Rightarrow 3sin^2x=1-sin^2x[/tex]

Finally, [tex]sinx=\pm\frac{1}{2} \Rightarrow x=\pi n\pm \frac{\pi}{6}[/tex] , n integral

I tested these values on a graphing calculator for [itex]0\leq x\leq 2\pi[/itex] because I didn't believe that there should be 4 max/min values, but only 2-3. I found that [itex]\pi/6[/itex] and [itex]7\pi/6[/itex] were max/min values but [itex]5\pi/6[/itex] and [itex]11\pi/6[/itex] were rather just random points on the function and its derivative.

So I'm guessing that I found extra wrong values when I squared both sides of the derivative? Is it possible to square the equation, but quickly find which values in the results aren't valid and need to be scrapped? After solving the equation this way, I realized that the values that are valid are located in the first and third quadrants which are both positive for the tangent function and thus I could've divided the derivative through by [itex]cosx[/itex].