1
sinx≠12/13⇒cosx≠5/13
cosx=t
26t²-23t+5=0
D=529-520=9
t1=(23-3)/52⇒cosx=5/13 не удов усл
t2=(23+3)/52=1/2⇒cosx=1/2⇒x=-π/3+2πk U x=π/3+2πk,k∈z
2
3^x=t
t²-31t+108≤0
t1+t2=31 U t1*t2=108
t1=4
t2=27
+ _ +
-----[4]--------------[27]-------------
4≤t≤27
4≤3^x≤27
log(3)4≤x≤3
x∈[log(3)4;3]