y1=arcsin(sinx)的定义域为R.值域为[-π/2,π/2]
y2=arccos(cosx)的定义域为R.值域为[0,π]
故
(a)当x∈[2kπ,2kπ+π/2](第一象限)时,y1=x-2kπ,y2=x-2kπ
y1+y2=π/2
2x-4kπ=π/2
x=2kπ+π/4 .(1)
(b)当x∈[2kπ+π/2,2kπ+π](第二象限)时,
y1=π-(x-2kπ),y2=x-2kπ
∴y1+y2=π/2
π-(x-2kπ)+x-2kπ=π/2
无解
(c)当x∈[2kπ+π,2kπ+3π/2](第三象限)时,
y1=-[x-(2k+1)π],y2=x-(2k+1)π
∴y1+y2=π/2
-[x-(2k+1)π]+x-(2k+1)π=π/2
无解
当x∈[2kπ+3π/2,2kπ+2π](第四象限)时,
y1=x-(2k+2)π,y2=x-(2k+1)π
∴y1+y2=π/2
x-(2k+2)π+x-(2k+1)π=π/2
2x-(4k+3)π=π/2
x=2kπ+7π/4 .(2)
综合(1),(2)可知,
原方程的解集为
{x|x=x=2kπ+π/4或x=2kπ+7π/4 (k为整数)}