32^n+2^n*8^n+2^4n= 2^5n+2^n*2^3n+2^4n=2^5n+2^4n+2^4n=2^4n(2^n+2)=2^5n+2^8n
Решение
Sin(3x/2 + π/12) < √2/2
- π - arcsin(√2/2) + 2πk < 3x/2 + π/12 < arcsin(√2/2) + 2πk, k ∈ Z
- π - π/4 + 2πk < 3x/2 + π/12 < π/4<span> + 2πk, k ∈ Z
</span>- 5π/4 + 2πk < 3x/2 < π/4<span> + 2πk, k ∈ Z
</span>- 5π/4 - π/12 + 2πk < 3x/2 < π/4 - π/12 + 2πk, k ∈ Z
- 4π/3 + 2πk < 3x/2 < π/6 <span>+ 2πk, k ∈ Z
</span>- 8π/3 + 4πk < 3x < π/3 <span>+ 4πk, k ∈ Z
</span>- 8π/9 + 4πk/3 < <span>x < </span>π/9 <span>+ 4πk/3, k ∈ Z</span>
Решение во вложении........................................................