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Given, $x,y \in [0,10]$ then find the no. of solutions of $3^{\sec^2 (x)-1}\sqrt{9y^2-6y+2}\leq 1$. I have tried it but got the answer to be infinite since $\sqrt{9y^2-6y+2}$ may also take negative values. Any help will be appreciated.

Chandramauli Chakraborty
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    As in the post-https://math.stackexchange.com/questions/2299199/square-root-of-x/2299209#2299207, I think you have doubt on the meaning of $\sqrt{x}$. To avoid confusion, $\sqrt{x}$ is defined as the non-negative square root of $x$ when $x\ge0$. – CY Aries May 27 '17 at 19:25
  • If we take both the positive and negative square roots, then $\sqrt{1}+\sqrt{4}$ could mean $1+2$, $1-2$, $-1+2$ and $-1-2$. It becomes so complicated. – CY Aries May 27 '17 at 19:27

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$\sqrt{9y^2-6y+2}$ can only take non-negative values.

Note that $9y^2-6y+2=(3y-1)^2+1\ge1$. So

$$3^{\tan^2x}=3^{\sec^2x-1}\le\frac{1}{\sqrt{9y^2-6y+2}}\le1$$

This implies that

$$\tan^2x\le0$$.

The only possibility is $\tan x=0$ and $3y-1=0$, i.e. $(x,y)=(0,\frac{1}{3})$, $(\pi,\frac{1}{3})$, $(2\pi,\frac{1}{3})$ or $(3\pi,\frac{1}{3})$.

CY Aries
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