8.118. $3(1-\sin t)+\sin ^{4} t=1+\cos ^{4} t$.

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8.118. $3(1-\sin t)+\sin ^{4} t=1+\cos ^{4} t$.
## Solution.

From the condition

$$
\begin{aligned}
& 3-3 \sin t+\sin ^{4} t-1-\left(\cos ^{2} t\right)^{2}=0, \sin ^{4} t-3 \sin t+2-\left(1-\sin ^{2} t\right)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2} t-3 \sin t+1=0
\end{aligned}
$$

Solving the last equation as a quadratic equation in terms of $\sin t$, we get $(\sin t)_{1}=\frac{1}{2}, t_{1}=(-1)^{k} \frac{\pi}{6}+\pi k, k \in Z$

$(\sin t)_{2}=1, t_{2}=\frac{\pi}{2}+2 \pi n=\frac{\pi}{2}(4 n+1), n \in Z$.

Answer: $\quad t_{1}=(-1)^{k} \frac{\pi}{6}+\pi k ; t_{2}=\frac{\pi}{2}(4 n+1), k, n \in Z$.