a,sin(πsin2x)=1
=> πsin2x = π/2+2kπ
=> sin2x=1/2+2k.
do |sin2x| ≤ 1 =>k=0 vì:
k≤-1 =>|1/2 + 2k| = -1/2-2k ≥ -1/2 - 2.(-1) = 3/2 >1
k≥ 1 =>|1/2 + 2k| = 1/2+2k ≥ 1/2 + 2.1 = 5/2 >1
=> sin2x = 1/2 =>2x=π/6+2kπ và 2x=5π/6+2kπ
=>x = π/12+kπ và x = 5π/12 + kπ.
b, tan[π/4.(cosx+sinx)] = 1 <=> π/4.(cosx+sinx) = π/4 + kπ
<=> cosx + sinx = 1+4k <=> √2.cos(x-π/4) = 1+4k <=> cos(x+π/4) = (1+4k)/√2có đk: -1 ≤ (1+4k)/√2 ≤ 1 <=> (-1-√2)/4 ≤ k ≤ (-1+√2)/4 => k = 0
=> cos(x-π/4) = 1/√2
=> cos(x-π/4) =cosπ/4
=>x-π/4=π/4+k2π=>x=π/2+k2π
và x-π/4=-π/4+k2π=>x=k2π