Solve the following inequality:

$$
\sin 4 x+\cos 4 x+\sin 5 x+\cos 5 x<0 \text {. }
$$
It is sufficient to find the values of $x$ satisfying the inequality in the interval

$$
0 \leq x<360^{\circ}
$$

because increasing $x$ by $360^{\circ}$ also increases $4 x$ and $5 x$ by multiples of $360^{\circ}$, thus the value of each term on the left side and the entire left side remains unchanged.

The expression $K$ on the left side is transformed by converting the sum of its $2-2$ terms into a product using the identity

$$
\begin{aligned}
\sin u+\cos u & =\sin u+\sin \left(90^{\circ}-u\right)=2 \sin 45^{\circ} \cos \left(u-45^{\circ}\right)= \\
& =\sqrt{2} \cos \left(u-45^{\circ}\right)
\end{aligned}
$$

and then using the identity for converting the sum of two cosines into a product:

$$
\begin{aligned}
K & =\sqrt{2} \cos \left(4 x-45^{\circ}\right)+\sqrt{2} \cos \left(5 x-45^{\circ}\right)= \\
& =2 \sqrt{2} \cos \left(\frac{9}{2} x-45^{\circ}\right) \cos \frac{1}{2} x
\end{aligned}
$$

Accordingly, (1) holds if and only if the variable factors in (3) have opposite signs.

Let's find all values of $x$ in (2) where one of the factors of $K$ is zero.

$$
\begin{gathered}
\cos \left(\frac{9}{2} x-45^{\circ}\right)=0, \text { if } \frac{9}{2} x-45^{\circ}=(2 k+1) 90^{\circ}, \text { i.e., } \\
x=30^{\circ}+k \cdot 40^{\circ} \\
\cos \frac{x}{2}=0, \quad \text { if } \frac{x}{2}=(2 k+1) 90^{\circ}, \quad x=180^{\circ}+k \cdot 360^{\circ}
\end{gathered}
$$

From the first factor, for $0 \leq k \leq 8$, and from the second factor, for $k=0$, we get values that fall within (2), so $K=0$ at the following points:

$$
30^{\circ}, \quad 70^{\circ}, 110^{\circ}, 150^{\circ}, 180^{\circ}, 190^{\circ}, 230^{\circ}, 270^{\circ}, 310^{\circ}, 350^{\circ} .
$$

At each of these points, the sign of one of the factors in (3) changes, and there is no value in (4) where the signs of both factors change simultaneously. Therefore, the sign of $K$ changes to the opposite at each value in (4).

At $x=0^{\circ}$, (1) does not hold because the left side is 2, which is positive. Therefore, the solution to (1) in the interval (2) is:

$$
\begin{gathered}
30^{\circ}<x<70^{\circ}, \quad 110^{\circ}<x<150^{\circ}, \quad 180^{\circ}<x<190^{\circ} \\
230^{\circ}<x<270^{\circ}, \quad 310^{\circ}<x<350^{\circ} .
\end{gathered}
$$

The graph shows the variation of $K$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b54d2ad0364d179f6cfag-1.jpg?height=924&width=1800&top_left_y=1690&top_left_x=152)