7. Find all values of $a$ for each of which exactly one of the following two statements is true:
1) «The equation $\cos (\cos x)+\sin (\sin x)=a$ has exactly two roots on the interval $[0 ; \pi]$ »
2) «The equation $\sin ^{4} x+\cos ^{4} x+\sin 2 x=a$ has roots».

Answer: $\left[-\frac{1}{2} ; \cos 1\right) \cup\left(\frac{3}{2} ; 1+\sin 1\right)$.
Solution. 1) The function $f(x)=\cos (\cos x)+\sin (\sin x)$ increases on the interval from 0 to $\frac{\pi}{2}$ (each of the terms is a monotonically increasing function) and decreases on the interval from $\frac{\pi}{2}$ to $\pi$ (since $f(\pi-x)=f(x)$). Therefore, $E(f)=[f(0) ; f(\pi / 2)]=[\cos 1 ; 1+\sin 1]$. This equation has exactly two roots on the interval $[0 ; \pi]$ when $a \in [\cos 1 ; 1+\sin 1)$.

2) In the second equation, we use the substitution $t=\sin 2 x$:

$$
\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x+\sin 2 x=a \Leftrightarrow 1-\frac{t^{2}}{2}+t=a \Leftrightarrow t^{2}-2 t-2=-2 a
$$

The range of the function $g(t)=t^{2}-2 t-2=(t-1)^{2}-3$ on the interval $[-1 ; 1]$ is the set $E(g)=[g(1) ; g(-1)]=[-3 ; 1]$. This equation has roots if and only if $-2 a \in[-3 ; 1]$, that is, $a \in\left[-\frac{1}{2} ; \frac{3}{2}\right]$.

3) Since $1+\sin 1>1+\sin \frac{\pi}{6}=\frac{3}{2}$, exactly one of the statements 1), 2) is true when $a \in\left[-\frac{1}{2} ; \cos 1\right) \cup\left(\frac{3}{2} ; 1+\sin 1\right)$.

Answer to option 2: $\left[-\frac{3}{2} ;-1+\cos 1\right) \cup\left(\frac{1}{2} ; \sin 1\right)$.

Answer to option 3: $\left(-1-\sin 1 ;-\frac{3}{2}\right] \cup\left(-\cos 1 ; \frac{1}{2}\right]$.

Answer to option 4: $\left(-\sin 1 ;-\frac{1}{2}\right) \cup\left(1-\cos 1 ; \frac{3}{2}\right]$.