1> 2sin2x – 5sinx.cosx – cos2x
= -2
Nếu
như: sinx = 0 thì cosx = 1 và ngược lại
Xét cos2x = 0 ó 2 =
-2 là vô lí, vậy cos2x ≠ 0
ó 2sin2x/
cos2x – 5sinx.cosx/ cos2x – cos2x/ cos2x
= -2/ cos2x
ó 2tan2x
– 5tanx – 1 = -2. (1 + tan2x)
ó 2tan2x
– 5tanx – 1 = -2 - 2tan2x
ó 2tan2x
– 5tanx – 1 + 2 + 2tan2x = 0
ó 4tan2x
– 5tanx + 1 = 0
Đặt
tanx = t
t1 = 1
t2 = ¼
·
t1 = 1 ó tanx
= 1
ó tanx
= π/4
ó x = π/4 + k π (k ∈ Z)
·
t2 = ¼ ó tanx
= ¼
ó tanx
= arc tan1/4
ó x = arc tan ¼ + k π (k ∈ Z)
2> 4sin2x + 3√3sin2x
– 2cos2x = 4
ó 4sin2x
+ 3√3.2sinx.cosx – 2cos2x = 4
Xét sin2x = 0 ó -2 =
4 là vô lí, vậy sin2x ≠ 0
ó 4sin2x/
sin2x + 6√3.sinx.cosx/ sin2x – 2cos2x / sin2x
= 4/ cos2x
ó 4 +
6√3cotx – 2cot2x = 4. (1 + cot2x)
ó 4 +
6√3cotx – 2cot2x – 4 – 4cot2x = 0
ó
-6cot2x + 6√3cotx = 0
ó
cotx. (-6cotx + 6√3) = 0
ó cotx
= 0
cotx = (-6√3/
6) = -√3
ó x = k π
x = π/6 +
k π (k
∈ Z)
3> 6sinx – 2cos3x = 5sin2x.cosx
ó
6sinx – 2cos3x = 5.2sinx.cosx.cosx
ó
6sinx – 2cos3x = 10.sinx.cos2x
Xét cos3x = 0 ó 6 =
0 là vô lí, vậy cos3x ≠ 0
ó
6sinx/ cos3x – 2cos3x/ cos3x = 10.sinx.cos2x/
cos3x
ó 6.
(1 +tan2x) – 2 = 10.tanx
ó 6 +
6tan2x – 2 – 10tanx =
0
ó 6tan2x
– 10tanx + 4 = 0
Đặt
tanx = t
t1 = 1
t2 = 2/3
·
t1 = 1 ó tanx
= 1
ó tanx
= π/4
ó x = π/4 + k π (k ∈ Z)
·
t2 = 2/3 ó tanx = 2/3
ó tanx
= arc tan2/3
ó x = arc tan2/3 + k π (k ∈ Z)
4> 2cos3x – sin3x =
cosx + sinx
Xét cos3x = 0 ó -1 =
1 là vô lí, vậy cos3x ≠ 0
ó 2cos3x/
cos3x – sin3x/ cos3x = cosx/ cos3x
+ sinx/ cos3x
ó 2 - tan3x = 1/ cos2x + sinx/cosx . 1/ cos2x
ó 2 -
tan3x = (1 +
tan2x) + tanx.(1 + tan2x)
ó 2 - tan3x = 1 + tan2x + tanx + tan3x
ó 2 –
tan3x – 1 – tan2x – tan – tan3x = 0
ó -2tan3x
– tan2x – tan + 1 = 0
Đặt
tanx = t
t1 = ½
·
t1 = ½ ó tanx
= ½
ó tanx = arc tan1/2
ó x = arc tan1/2 + k π (k ∈ Z)
5> sinx. 2sinx.cosx + sin3x =
6cos3x
ó 2sin2x.cosx
+ 3sinx - 4sin3x = 6cos3x
Xét sin3x = 0 ó 0 =
6 là vô lí, vậy sin3x ≠ 0 (lấy sin3x xét để có thể làm mẫu chung)
ó 2sin2x.cosx/
sin3x + 3sinx/sin3x – 4sin3x/sin3x = 6cos3x/sin3x
ó 2cotx +
3/sin2x - 4 = 6cot3x
ó 2cotx + 3.
(1 + cot2x) - 4 = 6cot3x
ó 2cotx + 3
+ 3cot2x - 4 = 6cot3x
ó -6cot3x + 3cot2x +
2cotx -1 =0
Đặt
cotx = t
t1 = -0.6
t2 = 0.6
t3 = ½
·
t1 = -0.6 ó cotx
= -0.6
ó cotx = arc cot-0.6
ó x
= arc cot-0.6 + k π (k ∈ Z)
·
t2 = 0.6 ó cotx = 0.6
ó cotx
= arc cot 0.6
ó x = arc cot 0.6 + k π (k ∈ Z)
·
t3
= ½ ó cotx = ½
ó cotx = arc cot ½
ó
x = arc cot ½ + k π (k ∈ Z)
1 nhận xét:
bài 3) 6sinx/ cos3x = 6. (1 +tan^2x) cho em hỏi ngu sao bằng vậy ạ
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