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How do we solve the following question?

If $x, y \in [0,10]$ then what is the number of solutions $(x, y)$ of the >inequation $3^{sec^2x - 1} \sqrt{9y^2-6y+2} \le 1$

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Hint:

$\sec^2x-1=\tan^2x\ge0$ for all real $x$

$\implies 3^{\sec^2x-1}\ge3^0$

and $(3y-1)^2+1\ge1$ for real $y$

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i would use that $$(\sec(x)^2-1)\ln(3)+\frac{1}{2}\ln(9y^2-6y+2)\le 0$$ after taking the logarithm on both sides

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HINT.-We have $$\sqrt{9y^2-6y+2}\ge1\text{ and just in }x=\frac13\text{ is equal to 1}$$

$$3^{\tan^2(x)}\ge1\text{ and is equal to 1 when } x=k\pi;\space k\in \mathbb Z$$

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