# Minimizing 4\sec^2(x)+9\csc^2(x) for x in the first qu

Minimizing $$\displaystyle{4}{{\sec}^{{2}}{\left({x}\right)}}+{9}{{\csc}^{{2}}{\left({x}\right)}}$$ for x in the first quadrant. Discrepancy in solution
Using derivatives, I am able to show that the minimum value of the expression-in-question is equal to 25. I also verified this with Desmos graphing app. However, when I tried doing this using basic algebra, the answer turns out to be 26. I do not know why this is happening, but I should be extremely grateful to you if you can point out the error.
$$\displaystyle{4}{{\sec}^{{2}}{x}}+{9}{{\csc}^{{2}}{x}}=$$
$$\displaystyle{\left({2}{\sec{{x}}}-{3}{\csc{{x}}}\right)}^{{2}}+{12}{\sec{{x}}}{\csc{{x}}}$$
The value of the above expression will be minimum when that expression inside parenthesis equals zero; and that happens when $$\displaystyle{\tan{{x}}}={\frac{{32}}{}}$$. Using this we can say that $$\displaystyle{\sec{{x}}}={\frac{{\sqrt{{{13}}}}}{{{2}}}}$$ and $$\displaystyle{\csc{{x}}}={\frac{{\sqrt{{{13}}}}}{{{3}}}}$$
If now we substitute these values in the above expression, answer turns out to be 26. I don't know why this is happening, but please help me find the error in this approach.

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lovagwb
Hint:
$$\displaystyle{2}{\sec{{x}}}-{3}{\csc{{x}}}={0}$$ may not give us the minimum value as $$\displaystyle{\sec{{x}}}{\csc{{x}}}$$ is not constant
Use
$$\displaystyle{4}{{\sec}^{{2}}{x}}+{9}{{\csc}^{{2}}{x}}={4}{\left({1}+{{\tan}^{{2}}{x}}\right)}+{9}{\left({1}+{{\cot}^{{2}}{x}}\right)}={13}+{\left({2}{\tan{{x}}}−{3}{\cot{{x}}}\right)}{2}+{2}\cdot{2}\cdot{3}$$
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Vasquez

Let the minimum value be a. Then:
$$\begin{array}{}4\sec^2x+9\csc^2x=a \Rightarrow 4\sin^2 x+9\cos^2 x=a\cos^2 x\sin^2 x \\\Rightarrow 4−4\cos^2x+9\cos^2x=a\cos^2x(1−\cos^2x) \\\Rightarrow −a\cos^4x+(a−5)\cos^2x−4=0 \end{array}$$
and since $$\triangle=0, (a-5)^2-4(-a)(-4)=0 \ or \ a^2-26a+25=0 \Rightarrow a=1,25$$. But since $$4\sec^2x+9\csc^2x \geq 9\csc^2 x=9\sin^2 x \geq \frac91$$, the minimum value must be 25.
To check if the minimum value is attained in the first quadrant, substitute a=25 into the quadratic equation above, which must result in a perfect square as $$\triangle=0$$

user_27qwe

substitute
$$\cos^2(x)=t$$
Having
$$\begin{array}{}\sec(x)=\frac{1}{\cos(x)} \\cosec(x)=\frac{1}{\sin(x)} \\\frac{4}{\cos^2 x}+\frac{9}{\sin^2 x} \to min, \ 0 <x<\pi \\\frac{4\sin^2 x}{\sin^2 x \cos^2 x}+\frac{9\cos^2 x}{\sin^2 x \cos^2 x} \to min \end{array}$$
substitute $$t=\cos^2(x),\sin^2(x)=1−t$$
$$\frac{4+5t}{(1-t)t} \to min$$
finding minimum by differentiation
$$\frac{d}{dt} \frac{4+5t}{(1-t)t}=0$$
$$\frac{5t^2+8t-4}{t^2(1-t)^2}=0$$
hence
$$5t^2+8t-4=0, \ t \ne 0, \ t \ne 1$$
solving we get $$t=\frac25, \ -2$$ Skipping verification that this value is minimum, we get
$$\cos^2 x=\frac25$$$$x=\arccos (\sqrt{\frac{2}{5}})$$
and minimum value
$$\frac{4+5t}{(1-t)t}=\frac{6}{\frac35 \times \frac25}=25$$