9-6/2^(tgx)=3/2*2^(2cos(x-π/4)/√2cosx)
a)2cos(x-π/4)/√2cosx=2/√2(cosxcosπ/4+
sinxsinπ/4)/cosx=2/√2*√2/2(1+tgx)=
1+tgx
9-6/2^(tgx)=3•2^(1+tgx)/2
3(3-2/2^(tgx))=3•2^(tgx)
2^(tgx)=t>0
3-2/t=t
3t-2-t^2=0
t^2-3t+2=0
D=9-8=1
t1=(3+1)/2=2
t2=(3-1)/2=1
2^tgx=2
tgx=1;x=π/4+πk
2^tgx=1;tgx=0;x=πk
-3π<π/4+πk<-3π/2<br>-13/4k1=-3;x=π/4-3π=-11π/4
k2=-2;x=π/4-2π=-7π/4
-3π<πk<-3π/2<br>-3k3=-2;x=-2π
ответ -11π/4;-7π/4;-2π