Решение
Sin(3x/2 + π/12) < √2/2<br>- π - arcsin(√2/2) + 2πk < 3x/2 + π/12 < arcsin(√2/2) + 2πk, k ∈ Z<br>- π - π/4 + 2πk < 3x/2 + π/12 < π/4<span> + 2πk, k ∈ Z
- 5π/4 + 2πk < 3x/2 < π/4<span> + 2πk, k ∈ Z
- 5π/4 - π/12 + 2πk < 3x/2 < π/4 - π/12 + 2πk, k ∈ Z<br>- 4π/3 + 2πk < 3x/2 < π/6 <span>+ 2πk, k ∈ Z
- 8π/3 + 4πk < 3x < π/3 <span>+ 4πk, k ∈ Z
- 8π/9 + 4πk/3 < <span>x < </span>π/9 + 4πk/3, k ∈ Z