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CHÖÔNGV

PHÖÔNG TRÌNH ÑOÁI XÖÙNG THEO SINX, COSX a ( sin x + cos x ) + b sin x cos x = c

(1)

Caù c h giaû i Ñaët t = sin x + cos x vôùi ñieàu kieän t ≤ 2 π⎞ π⎞ ⎛ ⎛ Thì t = 2 sin ⎜ x + ⎟ = 2 cos ⎜ x − ⎟ 4⎠ 4⎠ ⎝ ⎝ Ta coù : t 2 = 1 + 2 sin x cos x neân (1) thaønh b 2 t −1 = c 2 ⇔ bt 2 + 2at − b − 2c = 0 Giaû i (2) tìm ñöôïc t, roà i so vôù i ñieà u kieä n t ≤ 2 at +

(

)

π⎞ ⎛ 2 sin ⎜ x + ⎟ = t ta tìm ñöôï c x 4⎠ ⎝ Baø i 106 : Giaû i phöông trình sin x + sin2 x + cos3 x = 0 ( *)

giaû i phöông trình

(

)

(*) ⇔ sin x (1 + sin x ) + cos x 1 − sin2 x = 0

⇔ (1 + sin x ) = 0 hay sin x + cos x (1 − sin x ) = 0 ⎡sin x = −1 (1 ) ⇔⎢ ⎢⎣sin x + cos x − sin x cos x = 0 ( 2 ) π • (1) ⇔ x = − + k2π ( k ∈ Z ) 2 π⎞ ⎛ •Xeùt ( 2 ) : ñaët t = sin x + cos x = 2 cos ⎜ x − ⎟ 4⎠ ⎝ ñieàu kieän t ≤ 2 thì t 2 = 1 + 2 sin x cos x t2 − 1 Vaä y (2) thaø n h t − =0 2 ⇔ t 2 − 2t − 1 = 0

⎡t = 1 − 2 ⇔⎢ ⎢⎣ t = 1 + 2 ( loaïi ) π⎞ ⎛ Do ñoù ( 2 ) ⇔ 2 cos ⎜ x − ⎟ = 1 − 2 4⎠ ⎝


2 π⎞ ⎛ ⇔ cos ⎜ x − ⎟ = − 1 = cos ϕ vôùi 0 < ϕ < 2π 4⎠ 2 ⎝ π 2 = ±ϕ + h2π, h ∈ , vôùi cos ϕ = −1 4 2 π 2 ⇔ x = ± ϕ + h2π, h ∈ , vôùi cos ϕ = −1 4 2 ⇔ x−

3 sin 2x ( *) 2 3 ( *) ⇔ −1 + ( sin x + cos x )(1 − sin x cos x ) = sin 2x 2 π⎞ ⎛ Ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

Baø i 107 : Giaû i phöông trình −1 + sin 3 x + cos3 x =

Thì t2 = 1 + 2sin x cos x ⎛ t2 − 1 ⎞ 3 2 Vaä y (*) thaø n h : −1 + t ⎜⎜ 1 − t −1 ⎟⎟ = 2 2 ⎝ ⎠

(

(

)

(

⇔ −2 + t 3 − t 2 = 3 t 2 − 1

)

)

⇔ t 3 + 3t 2 − 3t − 1 = 0

(

)

⇔ ( t − 1) t 2 + 4t + 1 = 0 ⇔ t = 1 ∨ t = −2 + 3 ∨ t = −2 − 3 ( loaïi ) π⎞ π 1 ⎛ vôùi t = 1 thì sin ⎜ x + ⎟ = = sin 4⎠ 4 2 ⎝ π π π 3π ⇔ x + = = k2π ∨ x + = + k2π, k ∈ 4 4 4 4 π ⇔ x = k2π ∨ x = + k2π , k ∈ 2 π⎞ 3−2 ⎛ = sin ϕ vôù i t = 3 − 2 thì sin ⎜ x + ⎟ = 4⎠ 2 ⎝

⇔ x+

π π 3−2 = ϕ + m2π ∨ x + = π − ϕ + m2π, m ∈ , vôùi = sin ϕ 4 4 2

⇔ x =ϕ−

π 3π + m2π ∨ x = − ϕ + m2π, m ∈ , vôùi 4 4

Baø i 108 :Giaû i phöông trình

3−2 2

2 ( sin x + cos x ) = tgx + cot gx ( *)

⎧sin x ≠ 0 Ñieà u kieä n ⎨ ⇔ sin 2x ≠ 0 ⎩cos x ≠ 0 sin x cos x Luù c ñoù (*) ⇔ 2 ( sin x + cos x ) = + cos x sin x

= sin ϕ


sin2 x + cos2 x 1 ⇔ 2 ( sin x + cos x ) = = sin x cos x sin x cos x π⎞ ⎛ Ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝

Thì t 2 = 1 + 2 sin x cos x vôùi t ≤ 2 vaø t 2 ≠ 1

2 t −1 3 ⇔ 2t − 2t − 2 = 0 (Hieå n nhieâ n t = ±1 khoâ n g laø nghieä m ) (*) thaøn h

(

⇔ t− 2

2t =

)(

2

)

2t 2 + 2t + 2 = 0

⎡t = 2 ⇔⎢ 2 ⎢⎣ t + 2t + 1 = 0 ( voâ nghieäm ) π⎞ ⎛ Vaä y ( *) ⇔ 2 sin ⎜ x + ⎟ = 2 4⎠ ⎝ π⎞ ⎛ ⇔ sin ⎜ x + ⎟ = 1 4⎠ ⎝ π π ⇔ x + = + k2π, k ∈ 4 2 π ⇔ x = + k2π, k ∈ 4

Baø i 109 : Giaû i phöông trình 3 ( cot gx − cos x ) − 5 ( tgx − sin x ) = 2 ( *) Vôù i ñieà u kieä n sin 2x ≠ 0 , nhaâ n 2 veá phöông trình cho sinxcosx ≠ 0 thì : ( *) ⇔ 3 cos2 x (1 − sin x ) − 5 sin2 x (1 − cos x ) = 2 sin x cos x ⇔ 3 cos2 x (1 − sin x ) − 5 sin2 x (1 − cos x ) = 5 sin x cos x − 3 sin x cos x ⇔ 3 cos x ⎡⎣cos x (1 − sin x ) + sin x ⎤⎦ − 5 sin x ⎡⎣sin x (1 − cos x ) + cos x ⎤⎦ = 0

⇔ 3 cos x ( cos x − sin x cos x + sin x ) − 5 sin x ( sin x − sin x cos x + cos x ) = 0 ⎡sin x + cos x − sin x cos x = 0 (1) ⇔⎢ ( 2) ⎢⎣3 cos x − 5 sin x = 0 ( Ghi chuù : A.B + A.C = A.D ⇔ A = 0 hay B + C = D ) π⎞ ⎛ Giaû i (1) Ñaë t t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝ Thì t2 = 1 + 2sin x cos x vôù i ñieà u kieä n : t ≤ 2 vaø t ≠ ±1

(1) thaøn h : t −

t2 − 1 = 0 ⇔ t 2 − 2t − 1 = 0 2

(

)

⎡ t = 1 + 2 loaïi do t ≤ 2 ⇔⎢ ⎢ t = 1 − 2 ( nhaän so vôùi ñieàu kieän ) ⎣


π⎞ 1− 2 ⎛ Vaä y sin ⎜ x + ⎟ = = sin α ( 0 < α < 2π ) 4⎠ 2 ⎝ π π ⎡ ⎡ x k2 x + = α + π = α − + k2π ⎢ ⎢ 4 4 ⇔⎢ ⇔⎢ ⎢ x + π = π − α + k2π, k ∈ ⎢ x = 3π − α + k2π, k ∈ ⎢⎣ ⎢⎣ 4 4 3 ( 2) ⇔ tgx = = tgβ ⇔ x = β + hπ, h ∈ ( vôùi 0 < β < π ) 5 Baø i 110 : Giaû i phöông trình 3tg3 x − tgx +

3 (1 + sin x ) cos x 2

⎛π x⎞ = 8 cos2 ⎜ − ⎟ ( *) ⎝4 2⎠

Ñieà u kieä n : cos x ≠ 0 ⇔ sin x ≠ ±1

⎡ ⎛π ⎞⎤ Luù c ñoù : (*) ⇔ tgx 3tg 2 x − 1 + 3 (1 + sin x ) 1 + tg 2 x = 4 ⎢1 + cos ⎜ − x ⎟ ⎥ ⎝2 ⎠⎦ ⎣ = 4 (1 + sin x )

(

)

(

)

⇔ tgx ( 3tg2 x − 1) + (1 + sin x ) ⎡⎣3 (1 + tg2 x ) − 4 ⎤⎦ = 0 ⇔ ( 3tg 2 x − 1) ( tgx + 1 + sin x ) = 0

⇔ ( 3tg 2 x − 1) ( sin x + cos x + sin x cos x ) = 0 ⎡3tg 2 x = 1 (1) ⇔⎢ ⎢⎣sin x + cos x + sin x cos x = 0

(2)

π 1 3 ⇔ tgx = ± ⇔ x = ± + kπ 3 3 6 π⎞ ⎛ • Giaûi ( 2 ) ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2 vaø t ≠ ±1 •(1) ⇔ tg 2 x =

Thì t 2 = 1 + 2 sin x cos x t2 −1 (2) thaøn h : t + = 0 ⇔ t 2 + 2t − 1 = 0 2 ⎡ t = −1 − 2 loaïi do ñieàu kieän t ≤ 2 ⇔⎢ ⎢ t = −1 + 2 ( nhaän so vôùi ñieàu kieän ) ⎣

(

)

2 −1 π⎞ ⎛ Vaä y sin ⎜ x + ⎟ = = sin ϕ 4⎠ 2 ⎝ π π ⎡ ⎡ ⎢ x + 4 = ϕ + k2π, k ∈ ¢ ⎢ x = ϕ − 4 + k2 π, k ∈ ¢ ⇔⎢ ⇔⎢ ⎢ x + π = π − ϕ + k2π, k ∈ ¢ ⎢ x = 3π − ϕ + k2 π, k ∈ ¢ ⎣⎢ 4 ⎣⎢ 4


Baø i 111 : Giaû i phöông trình 2sin 3 x − sin x = 2 cos3 x − cos x + cos2x ( *)

(*) ⇔ 2 ( sin 3 x − cos3 x ) − ( sin x − cos x ) + sin 2 x − cos2 x = 0 ⇔ sin x − cos x = 0 hay 2 (1 + sin x cos x ) − 1 + ( sin x + cos x ) = 0 ⎡sin x − cos x = 0 (1) ⇔⎢ ⎢⎣sin x + cos x + sin 2x + 1 = 0 ( 2 ) • (1) ⇔ tgx = 1 ⇔x=

π + kπ, k ∈ ¢ 4

π⎞ ⎛ t = sin x + cos x = 2 cos x ⎜ x − ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n : t ≤ 2 •xeùt ( 2 ) ñaët

t 2 = 1 + sin 2x Vaäy ( 2 ) thaønh t + ( t 2 − 1) + 1 = 0

⇔ t ( t + 1) = 0 ⇔ t = 0 ∨ t = −1

π⎞ ⎛ Khi t = 0 thì cos ⎜ x − ⎟ = 0 4⎠ ⎝ π π ⇔ x − = ( 2k + 1) , k ∈ ¢ 4 2 3π ⇔x= + kπ, k ∈ ¢ 4 π⎞ 1 3π ⎛ Khi t = −1 thì cos ⎜ x − ⎟ = − = cos 4⎠ 4 2 ⎝ π 3π ⇔ x− =± + k2 π, k ∈ ¢ 4 4 π ⇔ x = π + k2 π hay x = − + k2 π, k ∈ ¢ 2

Baø i 112 : Giaû i phöông trình

sin x + sin 2 x + sin3 x + sin 4 x = cos x + cos2 x + cos3 x + cos4 x ( *)

Ta coù : (*) ⇔ ( sin x − cos x ) + ( sin 2 x − cos2 x ) + ( sin 3 x − cos3 x ) + ( sin 4 x − cos4 x ) = 0 ⇔ ( sin x − cos x ) = 0 hay 1 + ( sin x + cos x ) + (1 + sin x.cos x ) + ( sin x + cos x ) = 0 ⎡sin x − cos x = 0 (1) ⇔⎢ ⎢⎣2 ( sin x + cos x ) + sin x cos x + 2 = 0 ( 2 ) Ta coù : (1) ⇔ tgx = 1 π ⇔ x = + kπ, k ∈ ¢ 4


π⎞ ⎛ Xeù t (2) : ñaë t t = sin x + cos x = 2 cos ⎜ x − ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

Thì t 2 = 1 + 2 sin x cos x t2 −1 +2 = 0 (2) thaø n h 2t + 2 ⇔ t 2 + 4t + 3 = 0 ⇔ t = −1 ∨ t = −3 ( loaïi ) π⎞ 1 3π ⎛ khi t = -1 thì cos ⎜ x − ⎟ = − = cos 4⎠ 4 2 ⎝ π 3π ⎡ − = + k2 π, k ∈ ¢ x ⎢ 4 4 ⇔⎢ ⎢ x − π = − 3π + k2 π, k ∈ ¢ ⎢⎣ 4 4 ⎡ x = π + k2 π, k ∈ ¢ ⇔⎢ ⎢ x = − π + k2 π, k ∈ ¢ ⎣ 2

(

)

Baø i 113 : Giaû i phöông trình tg 2 x 1 − sin 3 x + cos3 x − 1 = 0 ( *)

Ñieà u kieä n : cos x ≠ 0 ⇔ sin x ≠ ±1 sin 2 x Luù c ñoù (*) ⇔ 1 − sin 3 x ) + cos3 x − 1 = 0 ( 2 cos x 2 ⇔ (1 − cos x )(1 − sin 3 x ) − (1 − cos3 x )(1 − sin 2 x ) = 0 ⇔ (1 − cos x )(1 − sin x ) = 0

hay (1 + cos x ) (1 + sin x + sin 2 x ) − (1 + cos x + cos2 x ) (1 + sin x ) = 0

⎡ cos x = 1 ( nhaän do ñieàu kieän ) ⎢ ⇔ ⎢sin x = 1 ( loaïi do ñieàu kieän ) ⎢ 2 2 2 2 ⎢⎣sin x + sin x cos x − cos x − sin x cos x = 0 ⎡ cos x = 1 ⇔⎢ 2 2 ⎣sin x − cos x + sin x cos x ( sin x − cos x ) = 0 ⎡ cos x = 1 ⇔⎢ ⎣sin x − cos x = 0 hay sin x + cos x + sin x cos x = 0 ⎡ cos x = 1 ∨ tgx = 1 ⇔⎢ ⎣sin x + cos x + sin x cos x = 0 ⎡ x = k2 π, k ∈ ¢ ⎢ π ⇔ ⎢ x = + kπ, k ∈ ¢ 4 ⎢ ⎢sin x + cos x + sin x cos x = 0 ⎣


xeù t pt sin x + cos x + sin x cos x = 0 ñaë t π⎞ ⎛ t = sin x + cos x = 2 cos x ⎜ x − ⎟ ñieàu kieän t ≤ 2 vaø t ≠ ±1 4⎠ ⎝ 2 ⇒ t = 1 + 2 sin x cos x t2 − 1 Ta ñöôï c phöông trình t + = 0 ⇔ t 2 + 2t − 1 = 0 2 ⎡ t = −1 − 2 ( loaïi ) ⇔⎢ ⎢⎣ t = − 1 + 2 ( nhaän so vôùi ñk ) π⎞ 2 −1 ⎛ = cos ϕ Vaä y cos ⎜ x − ⎟ = 4⎠ 2 ⎝ π π ⇔ x − = ±ϕ + k2 π, k ∈ ¢ ⇔ x = ± ϕ + k2 π, k ∈ ¢ 4 4

(

)

Baø i 114 : Cho phöông trình m ( sin x + cos x + 1) = 1 + sin 2x ( *) ⎡ π⎤ Tìm m ñeå phöông trình coù nghieäm thuoäc ñoaï n ⎢ 0, ⎥ ⎣ 2⎦ π⎞ ⎛ Ñaë t t = sin x + cos x = 2 sin ⎜ x − ⎟ , ñieà u kieä n t ≤ 2 4⎠ ⎝ 2 Thì t = 1 + sin 2 x Vaä y (*) thaø n h : m ( t + 1) = t 2 π π π 3π thì ≤ x + ≤ 2 4 4 4 2 π⎞ ⎛ ≤ sin ⎜ x + ⎟ ≤ 1 Do ñoù 2 4⎠ ⎝ ⇔1≤ t ≤ 2 ta coù m ( t + 1) = t 2

Neá u 0 ≤ x ≤

t2 (do t = -1 khoâ n g laø nghieä m cuû a phöông trình) t +1 t2 Xeù t y = treân ⎡⎣1, 2 ⎤⎦ t +1 t 2 + 2t Thì y ' = > 0 ∀t ∈ ⎡⎣1, 2 ⎤⎦ 2 ( t + 1) ⇔m=

Vaä y y taê n g treâ n ⎡⎣1, 2 ⎤⎦ ⎡ π⎤ Vaä y (*) coù nghieä m treâ n ⎢1, ⎥ ⇔ y (1) ≤ m ≤ y ⎣ 2⎦ 1 ⇔ ≤ m ≤ 2 2 −1 2

(

)

( 2)


Baø i 115 : Cho phöông trình cos3 x + sin 3 x = m sin x cos x ( *)

a/ Giaû i phöông trình khi m = 2 b/ Tìm m ñeå (*) coù nghieä m Ta coù : (*) ⇔ ( cos x + sin x )(1 − sin x cos x ) = m sin x cos x π⎞ ⎛ Ñaë t t = sin x + cos x = 2 cos x ⎜ x − ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

(

)

Thì t 2 = 1 + 2 sin x cos x ⎛ t2 − 1 ⎞ ⎛ t2 − 1 ⎞ = m Vaä y (*) thaø n h t ⎜ 1 − ⎟ ⎟ ⎜ 2 ⎠ ⎝ ⎝ 2 ⎠ ⇔ t ( 3 − t 2 ) = m ( t 2 − 1) a/ Khi m = 2 ta coù phöông trình t ( 3 − t 2 ) = 2 ( t 2 − 1)

(

)

⇔ t 3 + 2t 2 − 3t − 2 = 0

(

)(

)

⇔ t − 2 t 2 + 2 2t + 1 = 0 ⇔ t = 2 hay t = − 2 + 1 hay t = − 2 − 1( loaïi ) π⎞ π π ⎛ Vaä y • cos x ⎜ x − ⎟ = 1 ⇔ x − = k2 π, k ∈ ¢ ⇔ x = + k2 π, k ∈ ¢ 4⎠ 4 4 ⎝

π ⎞ 1− 2 ⎛ • cos ⎜ x − ⎟ = = cos α 4⎠ 2 ⎝ π π ⇔ x − = ±α + k2 π, k ∈ ¢ ⇔ x = ± α + k2π, k ∈ ¢ 4 4 2 2 b/ Xeù t phöông trình t ( 3 − t ) = k ( t − 1) ( **) Do t = ±1 khoâ n g laø nghieä m cuû a (**) neâ n 3t − t 3 * * ⇔ m = ( ) t2 − 1 3t − t 3 Xeù t y = 2 ( C ) treân ⎡⎣− 2, 2 ⎤⎦ \ {±1} t −1 −t 4 − 3 < 0∀t = ±1 Ta coù y ' = 2 2 t − 1 ( ) suy ra y giaûm treân ( −1,1 ) vaø lim y = + ∞ , lim− y = − ∞

x → − 1+

x→ 1

Do ñoù treân ( − 1,1 ) ⊂ ⎡⎣ − 2, 2 ⎤⎦ \ {±1} ta coù 3t − t 3 vôùi ∀m ∈ R (d) y = m caét (C) y = 2 t −1 Vaä y (*) coù nghieä m ∀m ∈ R


Baø i 116 : Cho phöông trình

1⎛ 1 1 ⎞ m ( sin x + cos x ) + 1 + ⎜ tgx + cot gx + + = 0 ( *) 2⎝ sin x cos x ⎟⎠ 1 a/ Giaû i phöông trình khi m = 2 ⎛ π⎞ b/ Tìm m ñeå (*) coù nghieä m treâ n ⎜ 0, ⎟ ⎝ 2⎠ Vôùi ñ ieàu kieän sin 2x ≠ 0 ta coù 1 ⎛ sin x cos x 1 1 ⎞ (*) ⇔ m ( sin x + cos x ) + 1 + ⎜ + + + =0 2 ⎝ cos x sin x sin x cos x ⎟⎠ ⇔ m sin 2x ( sin x + cos x ) + sin 2x + (1 + cos x + sin x ) = 0

⇔ m sin 2x ( sin x + cos x ) + sin 2x + 1 + cos x + sin x = 0 ⇔ m sin 2x ( sin x + cos x ) + ( sin x + cos x ) + sin x + cos x = 0 2

⎡sin x + cos x = 0 (1) ⇔⎢ ⎢⎣ m sin 2x + sin x + cos x + 1 = 0 ( 2 ) π⎞ ⎛ Xeù t (2) ñaë t t = sin x + cos x = 2 cos ⎜ x − ⎟ 4⎠ ⎝ 2 Thì t = 1 + sin 2 x Do sin 2x ≠ 0 neân t ≤ 2 vaø t = ±1

⎡t = 0 Vaä y (*) thaø n h : ⎢ 2 ⎢⎣ m ( t − 1) + t + 1 = 0 ⎡ t = 0 ( nhaän so ñieàu kieän ) ⇔⎢ ( do t ≠ −1) ⎢⎣ m ( t − 1) + 1 = 0 1 a/ Khi m = thì ta ñöôï c : 2 ⎡t = 0 ⎢ ⎢⎣ t = − 1 ( loaïi do ñieàu kieän ) Vaä y sinx + cosx = 0 ⇔ tgx = −1 π ⇔ x = − + kπ, k ∈ ¢ 4 π π π π b/ Ta coù : 0 < x < ⇔ − < x − < 2 4 4 4 Luù c ñoù 2 π⎞ ⎛ < cos ⎜ x − ⎟ ≤ 1 ⇒ 1 < t ≤ 2 2 4⎠ ⎝ Do t = 0 ∉ 1, 2 ⎤⎦

(


Neâ n ta xeù t phöôn g trình : m ( t − 1) + 1 = 0 ( **)

(**) ⇔ mt = m − 1 1 (do m = 0 thì (**) voâ nghieä m ) m 1 Do ñoù : yeâ u caà u baø i toaùn ⇔ 1 < 1 − ≤ 2 m ⎧ 1 ⎧m < 0 ⎪⎪− m > 0 ⎪ ⇔⎨ ⇔⎨ 1 ⎪1 − 2 ≤ 1 ⎪m ≤ 1 − 2 = − 2 − 1 ⎩ ⎪⎩ m

⇔ t = 1−

⇔ m ≤ − 2 −1 Baø i 117 : Cho f ( x ) = cos2 2x + 2 ( sin x + cos x ) − 3sin 2x + m 3

a/ Giaû i phöông trình f(x) = 0 khi m = -3 b/ Tính theo m giaù trò lôù n nhaá t vaø giaù trò nhoû nhaát cuûa f(x) 2

Tìm m cho ⎡⎣ f ( x ) ⎤⎦ ≤ 36 ∀x ∈ R

(

π⎞ ⎛ Ñaë t t = sin x + cos x = 2 cos ⎜ x − ⎟ ñieàu kieän t ≤ 2 4⎠ ⎝ 2 Thì t = 1 + sin 2 x

Vaø cos2 2x = 1 − sin 2 2x = 1 − ( t 2 − 1) = −t 4 + 2t 2 2

Vaä y f ( x ) thaønh g ( t ) = − t 4 + 2t 2 + 2t 3 − 3 ( t 2 − 1) + m a/ Khi m = -3 thì g(t) = 0 ⇔ −t 2 t 2 − 2t + 1 = 0

(

)

⇔ t = 0∨ t =1 vaäy khi m = -3 thì f( x) = 0 π⎞ 1 π⎞ ⎛ ⎛ ⇔ cos ⎜ x − ⎟ = 0 hay cos ⎜ x − ⎟ = 4⎠ 4⎠ 2 ⎝ ⎝ π π π π ⇔ x − = ( 2k + 1) hay x − = ± + k2π, k ∈ ¢ 4 2 4 4 3π π ⇔x= + kπ hay x = + k2 π ∨ x = k2 π, k ∈ ¢ 2 4 3 2 b/ Ta coù g ' ( t ) = −4t + 6t − 2t = −2t ( 2t 2 − 3t + 1) ⎧g ' ( t ) = 0 1 ⎪ Vaä y ⎨ ⇔ t = 0 ∨ t = 1∨ t = 2 ⎪⎩t ∈ ⎡⎣ − 2, 2 ⎤⎦ ⎛ 1 ⎞ 47 Ta coù : g ( 0 ) = 3 + m = g (1) , g⎜ ⎟ = +m ⎝ 2 ⎠ 16 g

( 2) = 4

2 − 3 + m,

g

( 2) = m −3−4

2

)


Vaä y : Maxf ( x ) =

Max g ( t ) = m + 3

t∈ ⎡⎣ − 2 , 2 ⎤⎦

x∈ ¡

Minf ( x ) = x∈ R

Min g ( t ) = m − 3 − 4 2

t ∈ ⎡⎣ − 2 , 2 ⎤⎦ 2

Do ñoù : ⎡⎣ f ( x ) ⎤⎦ ≤ 36, ∀x ∈ R ⇔ −6 ≤ f ( x ) ≤ 6, ∀x ∈ R ⎧Max f ( x ) ≤ 6 ⎪ ⇔⎨ R f (x) ≥ − 6 ⎪⎩Min R ⎧⎪m + 3 ≤ 6 ⇔⎨ ⎪⎩m − 3 − 4 2 ≥ −6 ⇔ 4 2 −3 ≤ m ≤ 3

(

)

2

Caù c h khaù c : Ta coù g ( t ) = −t 2 t 2 − 2t + 1 + 3 + m = − ⎡⎣ t ( t − 1) ⎤⎦ + 3 + m

Ñaë t u = t 2 − t

⎡ 1 ⎤ Khi t ∈ ⎡ − 2, 2 ⎤ thì u ∈ ⎢ − ,2 + 2 ⎥ = D ⎣ ⎦ ⎣ 4 ⎦ 2 Vaä y g ( t ) = h ( u ) = − u + 3 + m

Max f ( x ) = R

Min f ( x ) = R

Max g ( t ) = Max h ( u ) = m + 3 u∈D

t ∈ ⎡⎣ − 2 , 2 ⎤⎦

Min

t ∈ ⎣⎡ − 2 , 2 ⎦⎤

g ( t ) = Min h ( u ) = m − 3 − 4 2 u∈D

Chuù yù 1 : Phöông trình giaû ñoá i xöù n g a ( sin x − cos x ) + b ( sin x cos x ) = 0

ñaë t t = sinx – cosx π⎞ π⎞ ⎛ ⎛ thì t = 2 sin ⎜ x − ⎟ = − 2 cos ⎜ x + ⎟ 4⎠ 4⎠ ⎝ ⎝ 2 vôù i ñieà u kieä n t ≤ 2 thì t = 1 − 2 sin x cos x Baø i 118 : Giaû i phöông trình 2sin x + cot gx = 2sin 2x + 1 ( *)

Ñieà u kieä n : sin x ≠ 0 ⇔ cos x = ±1 cos x Luù c ñoù (*) ⇔ 2 sin x + = 4 sin x cos x + 1 sin x ⇔ 2 sin2 x + cos x = 4 sin2 x cos x + sin x

(

)

⇔ 2 sin2 x − sin x − cos x 4 sin2 x − 1 = 0 ⇔ sin x ( 2 sin x − 1) − cos x ( 2 sin x − 1) ( 2 sin x + 1) = 0 ⇔ 2 sin x − 1 = 0 hay sin x − cos x ( 2 sin x + 1) = 0 ⎡2 sin x − 1 = 0 ⇔⎢ ⎢⎣sin x − cos x − sin 2x = 0

(1 ) ( 2)


• Ta coù (1) ⇔ sin x = ⇔x=

1 ( nhaän do sin x ≠ 0) 2

π 5π + k2π ∨ x = + k2π, k ∈ 6 6

• Xeùt ( 2 ) Ñaët t = sin x − cos x =

π⎞ ⎛ 2 sin ⎜ x − ⎟ 4⎠ ⎝

Vôù i ñieà u kieä n t ≤ 2 vaø t ≠ ± 1 Thì t2 = 1 − sin 2x Vaä y (2) thaø n h : t − 1 − t 2 = 0

(

)

⇔ t2 + t − 1 = 0 −1 + 5 −1 − 5 ⇔t= ∨t= ( loaïi ) 2 2 π ⎞ −1 + 5 ⎛ Do ñoù : 2 sin ⎜ x − ⎟ = nhaän do t ≤ 2 vaø t ≠ ±1 4⎠ 2 ⎝ π⎞ 5 −1 ⎛ ⇔ sin ⎜ x − ⎟ = = sin ϕ 4⎠ 2 2 ⎝ π ⎡ ⎢ x − 4 = ϕ + k2π, k ∈ ⇔⎢ ⎢ x − π = π − ϕ + k2π, k ∈ ⎢⎣ 4 π ⎡ ⎢ x = ϕ + 4 + k2π, k ∈ ⇔⎢ ⎢ x = 5π − ϕ + k2π, k ∈ ⎢⎣ 4

(

)

Baø i 119 : Giaû i phöông trình cos 2x + 5 = 2 ( 2 − cos x )( sin x − cos x )( *)

(

)

Ta coù : ( *) ⇔ cos2 x − sin2 x + 5 = 2 ( 2 − cos x )( sin x − cos x ) ⇔ ( sin x − cos x ) ⎡⎣ 2 ( 2 − cos x ) + ( sin x + cos x ) ⎤⎦ − 5 = 0

⇔ ( sin x − cos x ) [sin x − cos x + 4] − 5 = 0 π⎞ ⎛ Ñaë t t = sin x − cos x = 2 sin ⎜ x − ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

(*) thaøn h : t ( t + 4 ) − 5 = 0 ⇔ t 2 + 4t − 5 = 0 ⇔ t = 1 ∨ t = −5 ( loaïi ) π⎞ 1 π ⎛ Vaä y ( *) ⇔ sin ⎜ x − ⎟ = = sin 4⎠ 4 2 ⎝


π π π 3π = + k2π ∨ x − = + k2π, k ∈ 4 4 4 4 π ⇔ x = + k2π ∨ x = π + k2π, k ∈ 2 ⇔ x−

Baø i 120 : Giaû i phöông trình cos3 x + sin 3 x = cos 2x ( *)

Ta coù (*) ⇔ ( cos x + sin x )(1 − sin x cos x ) = cos2 x − sin2 x ⇔ cos x + sin x = 0 hay 1 − sin x cos x = cosx − sin x ⎡sin x + cos x = 0 ⇔⎢ ⎢⎣sin x − cos x − sin x cos x + 1 = 0 Ta coù : (1) ⇔ tgx = −1 ⇔x=−

(1 ) ( 2)

π + kπ, k ∈ 4

π⎞ ⎛ Xeù t (2) ñaë t t = sin x − cos x = 2 sin ⎜ x − ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

Thì t2 = 1 − 2sin x cos x 1 − t2 (2) thaøn h t − + 1 = 0 ⇔ t 2 + 2t + 1 = 0 2 ⇔ t = −1 π⎞ 1 ⎛ ⎛ π⎞ vaä y (2) ⇔ sin ⎜ x − ⎟ = − = sin ⎜ − ⎟ 4⎠ 2 ⎝ ⎝ 4⎠ π π ⎡ x − = − + k2π, k ∈ ⎡ x = k2π, k ∈ ⎢ 4 4 ⇔⎢ ⇔⎢ ⎢ x = 3π + k2π, k ∈ 5 π π ⎢x − = + k2π, k ∈ ⎣ 2 ⎢⎣ 4 4 Baø i 121 : Cho phöông trình cos3 x − sin 3 x = m

(1 )

a/ Giaû i phöông trình (1) khi m = 1 baè n g caù c h ñaë t aå n phuï t = cos x − sin x ⎡ π π⎤ b/ Tìm m sao cho (1) coù ñuù n g hai nghieä m x ∈ ⎢ − , ⎥ ⎣ 4 4⎦ Ta coù (1) ⇔ ( cos x − sin x )(1 + sin x cos x ) = m π⎞ ⎛ Ñaë t t = cos x − sin x = 2 cos ⎜ x + ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n t ≤ 2

Thì t2 = 1 − 2sin x cos x ⎛ 1 − t2 ⎞ Vaä y (1) thaø n h : t ⎜⎜ 1 + ⎟=m 2 ⎟⎠ ⎝

(

)

⇔ t 3 − t 2 = 2m

( 2)


a/ Khi m = 1 thì (2) thaø nh t3 − 3t + 2 = 0 ⇔ ( t − 1) t 2 + t − 2 = 0

(

)

⇔ t = 1 ∨ t = −2 ( loaïi ) π⎞ 2 π π ⎛ Vaä y cos ⎜ x + ⎟ = ⇔ x + = ± + k2π, k ∈ 4⎠ 2 4 4 ⎝ π ⇔ x = k2π ∨ x = − + k2π, k ∈ 2 π π π π ⎡ ⎤ b/ Neá u x ∈ ⎢ − , ⎥ thì 0 ≤ x + ≤ 4 2 ⎣ 4 4⎦ π⎞ ⎛ neâ n 0 ≤ cos ⎜ x + ⎟ ≤ 1 4⎠ ⎝ π⎞ ⎛ ⇔ 0 ≤ t = 2 cos ⎜ x + ⎟ ≤ 2 4⎠ ⎝ nhaä n xeù t raè n g vôù i moãi t tìm ñöôï c treâ n ⎡⎣0, 2 ⎤⎦ ⎡ π π⎤ ta tìm duy nhaá t moä t x ∈ ⎢ − , ⎥ ⎣ 4 4⎦ 3 xeù t f ( t ) = −t + 3t treân ⎡⎣0, 2 ⎤⎦ ⇒ f ' ( t ) = −3t 2 + 3

⎡ π π⎤ vaä y (1) coù ñuù n g hai nghieä m x ∈ ⎢ − , ⎥ ⎣ 4 4⎦

⇔ ( d ) y = 2m caét ( C ) y = −t 3 + 3t treân ⎡⎣0, 2 ⎤⎦ taï i 2 ñieå m phaâ n bieä t ⇔ 2 ≤ 2m < 2 2 ⇔ ≤ m<1 2 Baø i 122 : Cho phöông trình 2 cos 2x + sin2 x cos x + sin x cos2 x = m ( sin x + cos x )( *)

a/ Giaû i phöông trình khi m = 2 ⎡ π⎤ b/ Tìm m ñeå phöông trình (*) coù ít nhaát moä t nghieä m treâ n ⎢0, ⎥ ⎣ 2⎦

Ta coù : ( * ) ⇔ 2 ( cos2 x − sin 2 x ) + sin x cos x ( sin x + cos x ) = m ( sin x + cos x )

⇔ cos x + sin x = 0

(1 ) hay 2 ( cos x − sin x ) + sin x cos x = m ( 2 )


π⎞ ⎛ Ñaë t t = cos x − sin x = 2 cos ⎜ x + ⎟ 4⎠ ⎝ 2 Thì t = 1 − 2 sin x cos x Ta coù : (1) ⇔ sin x = − cos x

(ñieà u kieä n t ≤ 2 )

π + kπ, k ∈ 4 1 − t2 Ta coù : (2) thaø nh 2t + =m 2 ⇔ −t 2 + 4t + 1 = 2m ( * *) ⇔ tgx = −1 ⇔ x = −

a/ Khi m = 2 thì (**) thaø n h t 2 − 4t + 3 = 0 ⇔ t = 1 ∨ t = 3 ( loaïi ) π⎞ 2 π π ⎛ ⇔ x + = ± + k2π, k ∈ vaä y cos ⎜ x + ⎟ = 4⎠ 2 4 4 ⎝ π ⇔ x = k2π ∨ x = − + kπ, k ∈ 2 Do ñoù : π π ( *) ⇔ x = − + kπ ∨ x = k2π ∨ x = − + k2π, k ∈ 4 2 π ⎡ π 3π ⎤ ⎡ π⎤ b/ Ta coù x ∈ ⎢0, ⎥ ⇔ x + ∈ ⎢ , ⎥ 4 ⎣4 4 ⎦ ⎣ 2⎦ 2 π⎞ 2 ⎛ vaä y − ≤ cos ⎜ x + ⎟ ≤ 2 4⎠ 2 ⎝ ⇒ −1 ≤ t ≤ 1 π ⎡ π⎤ Do nghieä m x = − + kπ ∉ ⎢0, ⎥ , ∀ k ∈ 4 ⎣ 2⎦ Neâ n yeâ u caà u baø i toaù n ⇔ ( * *) coù nghieä m treâ n [ −1,1]

Xeù t y = −t 2 + 4t + 1 thì y ' = −2t + 4 > 0 ∀t ∈ [ −1,1]

⇒ y taêng treân [ −1,1] Do ñoù : yeâ u caà u baø i toaùn ⇔ −4 = y ( −1) ≤ 2m ≤ y (1) = 4 ⇔ −2 ≤ m ≤ 2 * Chuù yù 2 : Phöông trình löôï n g giaù c daï n g a ( tgx ± cot gx ) + b ( tg 2 x + cot g 2 x ) = 0

ta ñaët t = tgx ± cot gx thì t 2 = tg 2 x + cot g 2 x ± 2 2 khi t = tgx + cot gx = thì t ≥ 2 ( do sin 2x ≤ 1) sin 2x Baø i 123 : Giaû i phöông trình 3tg 2 x + 4tgx + 4 cot gx + 3cot g 2 x + 2 = 0 ( *)


Ñaë t t = tgx + cot gx = Vôù i ñieà u kieä n t ≥ 2

2 sin 2x

Thì t 2 = tg 2 x + cot g 2 x + 2

(*) thaøn h : 3 ( t 2 − 2 ) + 4t + 2 = 0 ⇔ 3t 2 + 4t − 4 = 0 2 ⎡ t = ( loaïi do ñieàu kieän ) ⎢ ⇔ 3 ⎢ ⎣ t = −2

Ta coù : t = −2 ⇔

2 = −2 ⇔ sin 2x = −1 2 sin x

π + k2π, k ∈ 2 π ⇔ x = − + kπ, k ∈ 4 Baø i 124 : Giaû i phöông trình tgx + tg 2 x + tg 3 x + cotgx + cotg 2 x + cotg 3 x = 6 ( *) ⇔ 2x = −

Ta coù (*) ⇔ ( tgx + cot gx ) + ( tg 2 x + cot g 2 x ) + ( tg 3 x + cot g 3 x ) = 6

(

2

)

⇔ ( tgx + cot gx ) + ( tgx + cot gx ) − 2 + ( tgx + cot gx ) tg 2 x + cot g 2 x − 1 = 6 2 2 ⇔ ( tgx + cot gx ) + ( tgx + cot gx ) + ( tgx + cot gx ) ⎡( tgx + cot gx ) − 3⎤ = 8 ⎣ ⎦ 2 Ñaë t t = tgx + cot gx = ( ñieàu kieän t ≥ 2) sin 2x Vaä y (*) thaø n h : t + t 2 + t ( t 2 − 3) = 8

⇔ t 3 + t 2 − 2t − 8 = 0 ⎡t = 2 ⇔ ( t − 2 ) t 2 + 3t + 4 = 0 ⇔ ⎢ 2 ⎣ t + 3t + 4 = 0 ( voâ nghieäm ) ⇔t=2 2 Vaä y = 2 ⇔ sin 2x = 1 sin 2x π ⇔ 2x = + k2π, k ∈ 2 π ⇔ x = + kπ, k ∈ 4

(

)

Baø i 125 : Giaû i phöông trình 2 + 2tg 2 x + 5tgx + 5 cot gx + 4 = 0 ( *) 2 sin x Caù c h 1 : (*) ⇔ 2 1 + cot g 2 x + 2tg 2 x + 5 ( tgx + cot gx ) + 4 = 0

(

)


(

)

⇔ 2 tg 2 x + cot g 2 x + 5 ( tgx + cot gx ) + 6 = 0 2 ⇔ 2 ⎡( tgx + cot gx ) − 2⎤ + 5 ( tgx + cot gx ) + 6 = 0 ⎣ ⎦ 2 Ñaë t t = tgx + cot gx = , vôùi t ≥ 2 sin 2x Ta ñöôï c phöông trình : 2t 2 + 5t + 2 = 0 1 ⇔ t = −2 ∨ t = − ( loaïi ) 2 2 Vaä y ( *) ⇔ = −2 ⇔ sin 2x = −1 sin 2x π ⇔ 2x = − + k2π, k ∈ 2 π ⇔ x = − + kπ, k ∈ 4 Caù c h 2 : Ñaë t u = tgx (vôù i ñieà u kieä n u ≠ 0 )

Vaä y (*) thaø n h : 2 +

2 5 + 2u 2 + 5u + + 4 = 0 2 u u

⇔ 2 + 2u 4 + 5u 3 + 5u + 6u 2 = 0

(

)

⇔ ( u + 1) 2u 3 + 3u 2 + 3u + 2 = 0 ⇔ ( u + 1)

2

( 2u

2

)

+u+2 =0

⎡u = −1 ( nhaän ) ⇔⎢ 2 ⎢⎣2u + u + 2 = 0 ( voâ nghieäm ) Vaä y (*) ⇔ tgx = -1 π ⇔ x = − + kπ, k ∈ 4 Baø i 126 : Cho phöông trình 1 + cot g 2 x + m ( tgx + cot gx ) + 2 = 0 2 cos x 5 a/ Giaû i phöông trình khi m = 2 b/ Tìm m ñeå phöông trình coù nghieä m Ta coù : (1) ⇔ tg 2 x + cot g 2 x + m ( tgx + cot gx ) + 3 = 0 2 ( ñieàu kieän t ≥ 2) sin 2x ⇒ t 2 = tg 2 x + cot g 2 x + 2

Ñaë t t = tgx + cot gx =

Vaä y (1) thaø n h : t 2 + mt + 1 = 0 a/ Khi m =

( 2)

5 ta ñöôï c phöông trình 2t 2 + 5t + 2 = 0 2

(1 )


⇔ t = −2 ∨ t = −

1 ( loaïi ) 2

2 = −2 ⇔ sin 2x = −1 sin 2x π ⇔ 2x = − + k2π, k ∈ 2 π ⇔ x = − + kπ, k ∈ 4 b/ Caù c h 1 : Ta coù : (2) ⇔ mt = −1 − t 2 1 ⇔ m = − − t (do t = 0 khoâ n g laø nghieä m cuû a (2)) t 1 Xeù t y = − − t vôùi t ≥ 2 t 1 1 − t2 Thì y ' = 2 − 1 = t t2 Ta coù : y ' = 0 ⇔ t = ±1

Do ñoù

Do ñoù (1) coù nghieäm ⇔ (d) caét ( C ) treân ( −∞, −2] U [ 2, +∞ ) 5 5 ∨m≥ 2 2 5 ⇔ m ≥ 2 Caù c h 2 : Yeâ u caà u baø i toaù n ⇔ f ( t ) = t 2 + mt + 1 = 0 coù nghieä m t thoû a t ≥ 2 ⇔m≤−

Nhaä n xeù t raè n g do P = 1 neâ n neá u f(t) coù hai nghieäm t1 , t 2 ( vôùi t1 ≤ t2 ) vaø coù ⎪⎧ t1 ≤ 1 ⎧⎪ t1 ≥ 1 nghieäm thì ta coù ⎨ ∨⎨ ⎪⎩ t 2 ≥ 1 ⎪⎩ t 2 ≤ 1 Do ñoù : Yeâ u caà u baø i toaù n ⇔ t1 ≤ −2 < t1 < 2 ∨ −2 < t1 < 2 ≤ t 2

⎪⎧1f ( −2) ≤ 0 ⎪⎧1f ( 2 ) ≤ 0 ⇔⎨ ∨⎨ ⇔ ⎪⎩1f ( 2 ) > 0 ⎪⎩1f ( −2 ) > 0 5 5 ⇔m≥ ∨m≤− 2 2

⎧−2m + 5 ≤ 0 ⎧−2m + 5 > 0 ∨⎨ ⎨ ⎩2m + 5 > 0 ⎩2m + 5 ≤ 0


1.

BAØI TAÄP

Giaû i caù c phöông trình : a/ 1 + cos3 x − sin 3 x = sin x b/ cos3 x + cos2 x + 2 sin x − 2 = 0 c/ cos 2x + 5 = 2 ( 2 − cos x )( sin x − cos x ) d/ cot gx − tgx = sin x + cos x e/ sin 3 x − cos3 x = sin x − cos x f/ 1 + tgx = sin x + cos x π⎞ ⎛ g/ sin 2x + 2 sin ⎜ x − ⎟ = 1 4⎠ ⎝ k/ sin 2x − 12 ( sin x − cos x ) + 12 = 0

sin x + cos x =1 sin 2x + 1 1 − cos 2x 1 − cos3 x m/ = 1 + cos 2x 1 − sin3 x n/ 5 ( sin x + cos x ) + sin 3x − cos 3x = 2 2 ( 2 + sin 2x )

l/

o/ 1 + sin x + cos x + sin 2x + 2 cos 2x = 0 p/ sin 2 x cos x − cos 2x + sin x = cos2 x sin x + cos x r/ cos 2x + 5 = 2 ( 2 − cos x )( sin x − cos x )

2.

s/ cos2 x + sin 3 x + cos x = 0 t/ 4 sin3 x − 1 = 3sin x − 3 cos 3x Cho phöông trình sin 2x ( sin x + cos x ) = m (1) a/ Chöù n g minh neá u m > 2 thì (1) voâ nghieä m b/ Giaû i phöông trình khi m = 2

3.

Cho phöông trình sin 2x + 4 ( cos x − sin x ) = m

4.

a/ Giaû i phöông trình khi m = 4 b/ Tìm m ñeå phöông trình coù nghieä m Cho phöông trình : sin x cos x − m ( sin x + cos x ) + 1 = 0 a/ Giaû i phöông trình khi m = 2 b/ Tìm m ñeå phöông trình coù nghieä m

5.

( ÑS : m

≥ 1)

3 + 3tg 2 x = m ( tgx + cot gx ) = 1 2 sin x Tìm m ñeå phöông trình coù nghieä m ( ÑS : m ≥ 4 )

Cho phöông trình

Th.S Phạm Hồng Danh TT luyện thi ĐH CLC Vĩnh Viễn

Pt doixung theo sinx cosx  
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