Решите уравнение \({1+2\sin(3-2x)=0.}\) Выберите вариант ответа. \({(-1)^{n+2}\dfrac{\pi}{12}+1,5+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\) \({(-1)^{n+2}\dfrac{\pi}{12}+1,5+{\pi}n,n\in{\mathbf{Z}}}\) \({(-1)^{n+2}\dfrac{\pi}{12}-1,5+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\) \({(-1)^{n+2}\dfrac{\pi}{12}+3+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\)

Задание

Решите уравнение \({1+2\sin(3-2x)=0.}\) Выберите вариант ответа.

\({(-1)^{n+2}\dfrac{\pi}{12}+1,5+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\)
\({(-1)^{n+2}\dfrac{\pi}{12}+1,5+{\pi}n,n\in{\mathbf{Z}}}\)
\({(-1)^{n+2}\dfrac{\pi}{12}-1,5+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\)
\({(-1)^{n+2}\dfrac{\pi}{12}+3+\dfrac{{\pi}n}{2},n\in{\mathbf{Z}}}\)