Cosx-√3sinx=2(1/2cosx-√3/2sinx)=
2(cosπ/3*cosx-sinπ/3*sinx)=2cos(x+π/3)
cos(x+π/3)=(4-2a-4a²)/2=2-a-2a²
2-a-2a²€[-1;1]
{2-a-2a²≤1
{2-a-2a²≥-1
1)2-a-2a²-1≤0
2a²+a-1≥0
D=1+8=9=3²
a=(1±3)/4
a1=1;a2=-1/2
__+__-1/2__-__1__+__
a€(-бес;-1/2]+[1;+бес )
2)2-а-2а²+1≥0
2а²+а-3≤0
Д=1+24=25=5²
а=(-1±5)/4
а1=1;а2=-3/2
а€(-3/2;1)
ответ а€[-3/2;-1/2]
а=-3/2