cau 4
<=>sin(x+20°)=1/2=sin(30)
x1+20=30+360.k: x1=10+360k
x2+20=180-30+360.t=130+360t
0<x1<180;=>k=0
0<x2<180;=>t=0
x1+x2=10+130=140°
pA(D)

câu 5
sin x-3cosx =0
sinx =0 ko phai nghiem
<=>1-3cotx=0
x=actcot (1/3)+kπ
m=1/3
PA.(A)

2sin^2x +7sinx -4=0
<=>(sinx+4)(2sinx-1) =0
sinx+4≠0
<=>sinx=1/2=sin (π/6)
x1=π/6+k2π
x2=π-π/6+t2π
x1,x2>0; k,t ≥0;
x1 min =π/6
x2 min =5π/6
S min =π

cau 7
<=>
tan x=√3; &(cosx & 2cosx +1)≠0
tanx=√3
x=π/3+kπ
x≠π/2+tπ
x≠-π/3+nπ
k,t,n€Z
x≠π/2+tπ <=>k≠-π/6+t =>moi t€z
x≠2π/3+2nπ<=>
k≠2/3+n moi n €z
tong hop
x=π/3+kπ (k€z)
pa (A)

cau 8
2tan^2 x +3/cosx =-3
<=>2(tan^2+1)+3/(cosx)=-1
<=>2/cos^2 x +3/cosx +1=0
<=>(1/cosx +1)(2/cosx +1)=0
cosx =-1; x=(2k+1)π
cosx =2(vo nghiem)
tong hop
x=(2k+1)π (k€Z)
PA (B )

cau 20
<=>(sin^2-cos^2x)(sin^4+cos^4-sin^2xcos^2x)=-13/8(sin^2x-cos^2x )
sin^2 x=cos^2x
=> 4 diem
(sin^4+cos^4-sin^2xcos^2x)=-13/8
<=>(1-3sin^2xcos^2x )=-13/8
-3sin^2( 2x)/4=-13/8-1=-21/8
sin^2(2x)=(21.4)/(8.3)=7 /2>1 vo nghiem
tong hop
4 diem
PA (B)
 

Cho pt : cos2x+3cosx-m+2=0
Tìm m để pt có nghiệm x€ (0;pi trên 3)