1. Solve the equation $\sqrt{\sin ^{2} x+\lg ^{2} x-1}=\sin x+\lg x-1$.
Answer: $10, \pi / 2+2 n \pi, n-$ an integer, non-negative.

Instructions. By squaring and factoring, we get $\sin x=1$ or $\lg x=1$. From the first equation, we have $x=\pi / 2+2 n \pi, n-$ an integer. From the second, $x=10$. Since the logarithm is defined only for positive numbers, $n$ is non-negative. It is easy to verify that all the numbers indicated in the answer are suitable.

Criteria. Extra negative roots obtained: 3 points.