So when I graphed this equation I got a minimum of $x = 1.047$. How do I write this in terms of $\pi$, if I even did this right? Thanks! |
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Substitue $y=\tan x$, then $y$ takes values in $(0,\infty)$ and $\cot x=\frac{1}{y}$. Find the value of $y$ which makes $f=y+\frac{3}{y}$ minimal, and from there find $x$ |
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Since you’re finding an extremum, you (should) have $f\,'(x)=\sec^2x-3\csc^2x$. Setting this to $0$ and multiplying by $\sin^2x$ yields $\tan^2x-3=0$, or $\tan x=\pm\sqrt3$. (You’re permitted to multiply by $\sin^2x$ because it’s not $0$ at any point in the interval of interest.) Since $0<x<\frac{\pi}2$, you know that $\tan x>0$, so $\tan x=\sqrt 3$. This is a standard angle, one of the handful that you should simply know; what angle is it? |
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This function tends to $+\infty$ at the bounds, and it is continuous, so it reaches a global minimum where its derivative is zero. Now $$ f'(x)=\frac{1-4\cos^2x}{\sin^2x\cos^2x}. $$ Can you solve $f'(x)=0$ on $(0,\pi/2)$? Since there is a unique solution, this must be your minimum. |
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