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II. PHNG TRèNH BC HAI I VI MT HM S LNG GIC
I!"Z
2
sin sin : 0 1.t x hoaởc t x thỡ ủieu kieọ n t= =
Baứi 1. '$4?
B0
0
DGO2DGJC 0BH
0
D[H2D[JC
5BH2
O
DD[H
O
D2DJ
0
HD HB
( )
2
tan 1 3 tan 3 0x x+ =
OB
( )
2
4sin 2 3 1 sin 3 0x x + + =
\B
3
4cos 3 2 sin2 8cosx x x+ =
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Q1 +Z +(F>
2
sin 0asin x b x c+ + =
JD
1 1t
2
cos cos 0a x b x c+ + =
J2D
1 1t
2
tan tan 0a x b x c+ + =
JD
( )
2
x k k Z +
2
cot cot 0a x b x c+ + =
J2D
( )x k k Z
]B
0
DG2
0
DJ0 ^B2
0
0D[H20DG5JC
Baøi 2. '$4?
BH
0
5DG
( )
2 3 1 cos3 3x+ −
JH 0B20DG_2DGOJC
5BH2
0
A0[\DBG\2
0
A[5DBJ5 HB
( )
2
1
3 3 tan 3 3 0
cos
x
x
− + − + =
OB
3
cos x
G
0
DJ_ \B_[52DG
2
4
1 tan x+
JC
]B
2
1
sin x
J2DG5 ^B
2
1
cos x
G52
0
DJO
_B20D[52DJ
2
4cos
2
x
CB020DGDJ
4
5
Baøi 3. )2'$4?
sin3 cos3 3 cos2
sin
1 2sin2 5
x x x
x
x
+ +
+ =
÷
+
?>`'$4?
( )
0 ; 2
π
Baøi 4. )2'$4?2OD2DJ2HD20DG520DG?>`'$4?
( )
;−
π π
Baøi 5. '$4?
4 4 4
5
sin sin sin
4 4 4
x x x
+ + + − =
÷ ÷
π π
III. PHƯƠNG TRÌNH BẬC NHẤT THEO SINX VÀ COSX
DẠNG: a sinx + b cosx = c (1)
Cách 1:
• )!'$4?2
2 2
a b+
"'7
AB⇔
2 2 2 2 2 2
sin cos
a b c
x x
a b a b a b
+ =
+ + +
• +Z
( )
2 2 2 2
sin , cos 0, 2
a b
a b a b
= = ∈
+ +
α α α π
'$4?4a
2 2
sin .sin cos .cos
c
x x
a b
+ =
+
α α
2 2
cos( ) cos (2)
c
x
a b
⇔ − = =
+
α β
• +(F>"&'$4?%>,
2 2 2
2 2
1 .
c
a b c
a b
≤ ⇔ + ≥
+
• A0B
2 ( )x k k Z⇔ = ± + ∈
α β π
Cách 2:
E b
2
2 2
x
x k k= + ⇔ = +
π
π π π
%,>=FKc
E b
2 cos 0.
2
x
x k≠ + ⇔ ≠
π π
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+Z
2
2 2
2 1
tan , sin , cos ,
2
1 1
x t t
t thay x x
t t
−
= = =
+ +
"'7'$4? -d2
2
( ) 2 0 (3)b c t at c b+ − + − =
N?
2 0,x k b c≠ + ⇔ + ≠
π π
eA5B%>F
2 2 2 2 2 2
' ( ) 0 .a c b a b c= − − ≥ ⇔ + ≥
∆
A5B8/f>
C
8%'$4?
0
tan .
2
x
t=
Ghi chú:
E )0'g3R"& >,-
0E )23R=0?"(F>"&'$4?%>
2 2 2
.a b c+ ≥
5E *."hS*)
2 2 2 2 2 2
.sin .cos . sin cosy a x b x a b x x a b= + ≤ + + = +
2 2 2 2
sin cos
min max tan
x x a
y a b vaø y a b x
a b b
⇔ = − + = + ⇔ = ⇔ =
Baøi 1. '$4?
B
cos 3sin 2x x+ =
0B
6
sin cos
2
x x+ =
5B
3 cos3 sin3 2x x+ =
HB
sin cos 2 sin5x x x+ =
OB
( ) ( )
3 1 sin 3 1 cos 3 1 0x x− − + + − =
\B
3sin2 sin 2 1
2
x x
+ + =
÷
π
Baøi 2. '$4?
B
2
2sin 3 sin2 3x x+ =
0B
( )
sin8 cos6 3 sin6 cos8x x x x− = +
5B
3 1
8cos
sin cos
x
x x
= +
HB2D[
3sin 2cos
3
x x
= −
÷
π
OBODG2ODJ
2
25D \BA52D[HD[\B
0
G0J[5A52D[HD[\B
Baøi 3. '$4?
B5D[02DJ0 0B
3
2DGHD[
3
JC
5B2DGHDJ[ HB0D[O2DJO
Baøi 4. '$4?
B0
4
x
+
÷
π
G
4
x
−
÷
π
J
3 2
2
0B
3 cos2 sin2 2sin 2 2 2
6
x x x
+ + − =
÷
π
Baøi 5. ?"&'$4?AG0BDG2DJ0%>
Baøi 6. ?"&'$4?A0[BDGA[B2DJ[5K>
IV. PHƯƠNG TRÌNH ĐẲNG CẤP BẬC HAI
DẠNG: a sin
2
x + b sinx.cosx + c cos
2
x = d (1)
Cách 1:
• i&42DJC%2M=FKc
@*A2DJC
2
sin 1 sin 1.
2
x k x x⇔ = + ⇔ = ⇔ =±
π
π
• i
cos 0x ≠
8!'$4?AB2
2
cos 0x ≠
"'7
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2 2
.tan .tan (1 tan )a x b x c d x+ + = +
• +ZJD8"'('$4? -d2
2
( ) . 0a d t b t c d− + + − =
Cách 2:QRKS1 -
1 cos2 sin2 1 cos2
(1) . . .
2 2 2
x x x
a b c d
− +
⇔ + + =
.sin2 ( ).cos2 2b x c a x d a c⇔ + − = − −
A"j=,'$4? -."/0D
20DB
Baøi 1. '$4?
B
( ) ( )
2 2
2sin 1 3 sin .cos 1 3 cos 1x x x x+ − + − =
0B
( )
2 2
3sin 8sin .cos 8 3 9 cos 0x x x x+ + − =
5B
2 2
4sin 3 3sin .cos 2cos 4x x x x+ − =
HB
2 2
1
sin sin2 2cos
2
x x x+ − =
OB
( ) ( )
2 2
2sin 3 3 sin .cos 3 1 cos 1x x x x+ + − = −
\B
2 2
5sin 2 3sin .cos 3cos 2x x x x+ + =
]B
2 2
3sin 8sin .cos 4cos 0x x x x+ + =
^B
( ) ( )
2 2
2 1 sin sin2 2 1 cos 2x x x− + + + =
_B
( ) ( )
2 2
3 1 sin 2 3 sin .cos 3 1 cos 0x x x x+ − + − =
CB
4 2 2 4
3cos 4sin cos sin 0x x x x− + =
B2
0
DG5
0
DG
2 3
D2D[JC
0B02
0
D[5D2DG
0
DJC
Baøi 2. '$4?
B
5
DG0
0
D2
0
D[52
5
DJC 0B
2
2 1
3sin .cos sin
2
x x x
−
− =
Baøi 3. ?"&'$4?AGB
0
D[0DG02
0
DJ%>
Baøi 4. ?"&'$4?A5[0B
0
D[AO[0B0DG5A0GB2
0
DJCK >
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