Question

求 sinx=cosx/2 在[-π,π]上的解集

Answer 1

sinx=cosx/2,

2sin(x/2)*cos(x/2)-cos(x/2)=0,

cos(x/2)=0,或sin(x/2)=1/2,

因为:X属于[-π,π],

-π/2≤(x/2)≤π/2.

cos(x/2)=0,

则有,x/2=-π/2,或x/2=π/2,

x1=-π,x2=π,

sin(x/2)=1/2,-π/2≤(x/2)≤π/2.

x/2=π/6,

x3=π/3.

即,sinx=cosx/2 在[-π,π]上的解集是:

x1=-π,x2=π,x3=π/3.