# Problem 2. Maximum 14 points

Solve the equation:

$$
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\log _{\left(\operatorname{tg}^{4} x+1\right)^{2}} \sqrt{2}=\log _{16} \frac{\operatorname{ctg}^{4} x}{\operatorname{ctg}^{4} x+1}
$$
Solution

Let's carry out a series of simple transformations

$$
\begin{gathered}
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{2} \cdot \frac{1}{2} \cdot \log _{\operatorname{tg}^{4} x+1} 2-\frac{1}{4} \cdot \log _{2} \frac{\operatorname{ctg}^{4} x}{\operatorname{ctg}^{4} x+1}=0 \\
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{4} \cdot \log _{\operatorname{tg}^{4} x+1} 2-\frac{1}{4} \cdot \log _{2} \frac{\operatorname{ctg}^{4} x \cdot \operatorname{tg}^{4} x}{\left(\operatorname{ctg}^{4} x+1\right) \cdot \operatorname{tg}^{4} x}=0 \\
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{4} \cdot \log _{\operatorname{tg}^{4} x+1} 2-\frac{1}{4} \cdot \log _{2} \frac{1}{\operatorname{tg}^{4} x+1}=0 \\
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{4} \cdot\left(\log _{\mathrm{tg}^{4} x+1} 2+\log _{2}\left(\operatorname{tg}^{4} x+1\right)\right)=0
\end{gathered}
$$

Notice that the numbers $\log _{\operatorname{tg}^{4} x+1} 2, \log _{2}\left(\operatorname{tg}^{4} x+1\right)$ are positive and reciprocals of each other, and thus, as is known,

$$
\log _{\mathrm{tg}^{4} x+1} 2+\log _{2}\left(\operatorname{tg}^{4} x+1\right) \geq 2
$$

with equality holding only when $\log _{2}\left(\operatorname{tg}^{4} x+1\right)=1$, which is equivalent to $\operatorname{tg} x= \pm 1$ or $x= \pm \frac{\pi}{4}+\pi n, n \in \mathbb{Z}$.

Now let's consider the expression:

$$
4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}
$$

We will estimate it by applying the inequality between the geometric mean and the arithmetic mean (Cauchy's inequality) for positive numbers $a, b$:

$$
\frac{a+b}{2} \geq \sqrt{a b} \Leftrightarrow a+b \geq 2 \sqrt{a b}
$$

which turns into equality only when $a=b$.

We have:

$$
\begin{gathered}
4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x} \geq 2 \cdot \sqrt{4^{\sqrt{2} \sin x} \cdot 4^{\sqrt{2} \cos x}}=2^{\sqrt{2}(\sin x+\cos x)+1}= \\
=2^{2 \sin \left(x+\frac{\pi}{4}\right)+1} \geq 2^{-2+1}=\frac{1}{2^{\prime}}
\end{gathered}
$$

where we used the auxiliary argument formula:

$$
\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)
$$

as well as the boundedness of the sine: $\sin \left(x+\frac{\pi}{4}\right) \geq-1$. Note that equality in the obtained inequality holds only when equality is achieved in each of the two applied estimates, i.e., when:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 4 ^ { \sqrt { 2 } \operatorname { s i n } x } = 4 ^ { \sqrt { 2 } \operatorname { c o s } x } , } \\
{ \operatorname { s i n } ( x + \frac { \pi } { 4 } ) = - 1 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ \operatorname { s i n } x = \operatorname { c o s } x } \\
{ x + \frac { \pi } { 4 } = \frac { 3 \pi } { 2 } + 2 \pi k , k \in \mathbb { Z } , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
\operatorname{tg} x=1 \\
x=\frac{5 \pi}{4}+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.\right. \\
& \Leftrightarrow x=\frac{5 \pi}{4}+2 \pi k, k \in \mathbb{Z}
\end{aligned}
$$

Thus, the following estimate holds:

$$
\begin{gathered}
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{4} \cdot\left(\log _{\operatorname{tg}^{4} x+1} 2+\log _{2}\left(\operatorname{tg}^{4} x+1\right)\right) \geq \\
\geq \log _{4} \frac{1}{2}+\frac{1}{4} \cdot 2=-\frac{1}{2}+\frac{1}{2}=0
\end{gathered}
$$

Therefore, equality to zero is possible only when all applied estimates turn into equalities, or, equivalently, when the conditions of the system are satisfied:

$$
\left\{\begin{array}{l}
x= \pm \frac{\pi}{4}+\pi n, n \in \mathbb{Z} \\
x=\frac{5 \pi}{4}+2 \pi k, k \in \mathbb{Z}
\end{array}\right.
$$

From which we obtain

Answer: $x=\frac{5 \pi}{4}+2 \pi k, k \in \mathbb{Z}$.

## Evaluation Criteria