Verified answer

P= 3.sinx +√3.cosx

P\√3 = √3sinx + cosx

P\√3.cospi\3 =cos(x-pi\3)

P= 2.√3.cos(x-pi\3)

=> P max = 2.√3 khi cos(x-pi\3) =1

=> p max = 2.√3 khi x= kpi +pi\3

c1/ P=3sina +can3 cosa

=2can3(sina.can3/2+cosa.1/2)

=2can3(sina.sinpi/3+cosa.cospi/3)

=2can3(cos(a-pi/3))

pmax khi cos(a-pi/3)=1=cos o

vay a=pi/3 do a la goc nhon

c2/ ap dung bat dang thuc bunhiacopxky:

(ab+cd)<=can(a^2+c^2)(b^2+d^2) ta co:

P=3sina +can3 cosa<=can(9+3)(sin^2 +có^2)=can12=2can3

pmax=2can3

xay ra khi a/c=b/d=sin/cos=can3=tanpi/3

vay a=pi/3

chuc ban hoc tot!