8. Let param1. What is the largest possible value of param2?

It is given that param1. Find the largest possible value of param2.

| param 1 | param 2 | Answer |
| :---: | :---: | :---: |
| $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=3$ | $6 \sin x$ | 4 |
| $\frac{25 \sin ^{2} x-37+40 \cos x}{35-25 \cos ^{2} x-30 \sin x}=4$ | $10 \cos x$ | 8 |


| $\frac{33-16 \cos ^{2} x-24 \sin x}{16 \sin ^{2} x-19-8 \sqrt{7} \cos x}=2$ | $12 \sin x$ | 9 |
| :---: | :---: | :---: |
| $\frac{15-9 \cos ^{2} x+6 \sin x}{9 \sin ^{2} x-7+12 \sqrt{2} \cos x}=\frac{1}{2}$ |  |  |

## Condition

Let $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=3$. What is the largest possible value of $6 \sin x$?

Answer.

4

#
# Solution

Notice that $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=\frac{6-(3 \sin x-2)^{2}}{2+(3 \cos x+\sqrt{5})^{2}}$. The obtained expression can equal 3 only if $3 \sin x-2=0$ and $3 \cos x+\sqrt{5}=0$. Therefore, the expression $6 \sin x$ can only take the value 4.