Bài 5 trang 37 sgk giải tích 11
Giải các phương trình sau:
a) (cosx – sqrt3sinx = sqrt2);
b) (3sin3x – 4cos3x = 5);
c) (2sin2x + 2cos2x – sqrt2 = 0);
d) (5cos2x + 12sin2x -13 = 0).
Giải
a) (cosx – sqrt3sinx = sqrt2)
( Leftrightarrow {1 over 2}cos x – {{sqrt 3 } over 2}{mathop{rm sinx}nolimits} = {{sqrt 2 } over 2})
( Leftrightarrow cos x.cos {pi over 3} – sin xsin {pi over 3} = cos {pi over 4})
( Leftrightarrow cos left( {x + {pi over 3}} right) = cos {pi over 4})
( Leftrightarrow left[ matrix{
x + {pi over 3} = {pi over 4} + k2pi hfill cr
x + {pi over 3} = – {pi over 4} + k2pi hfill cr} right.)
( Leftrightarrow left[ matrix{
x = – {pi over {12}} + k2pi hfill cr
x = – {{7pi } over {12}} + k2pi hfill cr} right.(k inmathbb{Z} ))
b) (3sin3x – 4cos3x = 5)
( Leftrightarrow {3 over 5}sin 3x – {4 over 5}cos 3x = 1)
Đặt (alpha =arccos{3over5}) thì phương trình trở thành
(cosαsin3x – sinαcos3x = 1)( ⇔ sin(3x – α) = 1)
( ⇔ 3x – α = {piover2} + k2π)
( Leftrightarrow x = {pi over 6} + {alpha over 3} + {{k2pi } over 3}(k in mathbb{Z}))
c) (2sin2x + 2cos2x – sqrt2 = 0)
(Leftrightarrow {1 over {sqrt 2 }}sin 2x + {1 over {sqrt 2 }}cos 2x = {1 over 2})
( Leftrightarrow sin 2x.cos {pi over 4} + cos 2x.sin {pi over 4} = sin {pi over 6})
( Leftrightarrow sin left( {2x + {pi over 4}} right) = sin {pi over 6})
( Leftrightarrow left[ matrix{
2x + {pi over 4} = {pi over 6} + k2pi hfill cr
2x + {pi over 4} = pi – {pi over 6} + k2pi hfill cr} right.)
( Leftrightarrow left[ matrix{
x = – {pi over {12}} + kpi hfill cr
x = {{7pi } over {12}} + kpi hfill cr} right.(k in mathbb{Z}))
d) (5cos2x + 12sin2x -13 = 0)
( Leftrightarrow {5 over {13}}cos 2x + {{12} over {13}}sin 2x = 1)
Đặt (alpha = arccos{5over13}) thì phương trình trở thành
(cosαcos2x + sinαsin2x = 1 ⇔ cos(2x – α) = 1)
(⇔ 2x-alpha = k2π) (Leftrightarrow x={alphaover2}+kpi), ((k ∈ mathbb{Z}))
(trong đó (α = arccos{5over13})).
Bài 6 trang 37 sgk giải tích 11
Giải các phương trình sau:
a. (tan (2x + 1)tan (3x – 1) = 1);
b. (tan x + tan left( {x + {pi over 4}} right) = 1)
Lời giải:
a) (tan(2x + 1)tan(3x – 1) = 1)
(tan (2x + 1) = {1 over {tan (3x – 1)}})
(Leftrightarrow tan (2x + 1) = cot (3x – 1))
( Leftrightarrow tan (2x + 1) = tan left( {{pi over 2} – 3x + 1} right))
( Leftrightarrow 2x + 1 = {pi over 2} – 3x + 1 + kpi )
( Leftrightarrow x = {pi over {10}} + {{kpi } over 5}(k inmathbb{Z} )).
b) (tan x + tan left( {x + {pi over 4}} right) = 1)
(eqalign{
& Leftrightarrow tan x + {{tan x + tan {pi over 4}} over {1 – tan x.tan {pi over 4}}} = 1 cr
& Leftrightarrow tan x + {{tan x + 1} over {1 – tan x}} = 1 cr} )
Đặt (t = tan x), (điều kiện (t ≠ 1))phương trình trở thành
(t + frac{t+1}{1-t})= 1
(Leftrightarrow – {t^2} + 3t = 0 Leftrightarrow left[ matrix{
t = 0 hfill cr
t = 3 hfill cr} right.text{(thỏa mãn)})
( Leftrightarrow left[ matrix{
tan x = 0 hfill cr
tan x = 3 hfill cr} right. Leftrightarrow left[ matrix{
x = kpi hfill cr
x = arctan 3 + kpi hfill cr} right.(k in mathbb{Z}))
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