# 20588562-Chuong8-Phuong-Trinh-Luong-Giac-Khong-Mau-Muc

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```					PHÖÔNG TRÌNH LÖÔÏ N G GIAÙC KHOÂNG MAÃU MÖÏC
Tröôø n g hôï p 1: TOÅ N G HAI SOÁ KHOÂ N G AÂ M

CHÖÔNG VIII

AÙ p duïn g
Baø i 156

Neá u ⎨

⎧A ≥ 0 ∧ B ≥ 0 thì A = B = 0 ⎩A + B = 0

Giaû i phöông trình: 4 cos2 x + 3tg 2 x − 4 3 cos x + 2 3tgx + 4 = 0 (*)

Ta coù :

(*) ⇔ 2 cos x − 3

(

) +(
2

3tgx + 1

)

2

=0

⎧ 3 ⎪cos x = ⎪ 2 ⇔⎨ ⎪tgx = − 1 ⎪ 3 ⎩ π ⎧ x = ± + k2π, k ∈ ⎪ 6 ⎪ ⇔⎨ ⎪tgx = − 1 ⎪ 3 ⎩ ⇔x=−
Baø i 157

π + k2π, k ∈ 6

Giaû i phöông trình: 8 cos 4x.cos2 2x + 1 − cos 3x + 1 = 0 ( *)

Ta coù : ( *) ⇔ 4 cos 4x (1 + cos 4x ) + 1 + 1 − cos 3x = 0
⇔ ( 4 cos2 4x + 4 cos 4x + 1) + 1 − cos 3x = 0 ⇔ ( 2 cos 4x + 1) + 1 − cos 3x = 0 1 ⎧ ⎪cos 4x = − ⇔⎨ 2⇔ ⎪cos 3x = 1 ⎩ 1 ⎧ ⎪cos 4x = − 2 ⎪ ⇔⎨ ⎪ x = k2π , k ∈ ⎪ 3 ⎩ 1 ⎧ ⎪cos 4x = − 2 ⎨ ⎪3x = k2π, k ∈ ⎩
2

(coù 3 ñaàu ngoïn cung)

1 ⎧ cos 4x = − ⎪ ⎪ 2 ⇔⎨ 2π 2π ⎪x = − +m2π hay x = m2π hay x = + m2π , m ∈ ⎪ 3 3 ⎩ 2π ⇔x=± + m2π, m ∈ 3 (ta nhaä n k = ±1 vaø loaï i k = 0 )
Baø i 158 Giaû i phöông trình: sin 2 3x sin2 x + ( cos 3x sin3 x + sin 3x cos3 x ) = sin x sin2 3x ( *) 3sin 4x Ta coù : cos 3x.sin 3 3x + sin 3x.cos3 x = ( 4 cos3 x − 3 cos x ) sin 3 x + ( 3 sin x − 4 sin3 x ) cos3 x

= −3 cos x sin 3 x + 3 sin x cos3 x = 3 sin x cos x ( cos2 x − sin 2 x ) = 3 3 sin 2x. cos 2x = sin 4x 2 4 1 Vaäy: ( *) ⇔ sin 2 x + sin2 3x = sin x sin2 3x vaø sin 4x ≠ 0 4
1 1 ⎛1 ⎞ ⇔ ⎜ sin 2 3x − sin x ⎟ − sin4 3x + sin2 3x = 0 vaø sin 4x ≠ 0 4 4 ⎝2 ⎠ 1 ⎛1 ⎞ ⇔ ⎜ sin 2 3x − sin x ⎟ + sin 2 3x (1 − sin2 3x ) = 0 vaø sin 4x ≠ 0 4 ⎝2 ⎠ 1 ⎛1 ⎞ ⇔ ⎜ sin2 3x − sin x ⎟ + sin2 6x = 0 vaø sin 4x ≠ 0 16 ⎝2 ⎠ ⎧sin 4x ≠ 0 ⎪1 ⎪ ⇔ ⎨ sin 2 3x = sin x ⎪2 ⎪sin 3x = 0 ∨ cos 3x = 0 ⎩
2 2 2

⎧sin 4x ≠ 0 ⎧sin 4x ≠ 0 ⎪ ⎪ ⎪1 ⇔ ⎨sin 3x = 0 ∨ ⎨ = sin x ⎪sin x = 0 (VN) ⎪ 2 ⎩ ⎪sin 3x = ±1 ⎩ ⎧sin 4x ≠ 0 ⎪ 1 ⎪ ⇔ ⎨sin x = 2 ⎪ ⎪3 sin x − 4 sin 3 x = ±1 ⎩

⎧sin 4x ≠ 0 ⎪ ⇔⎨ 1 ⎪sin x = 2 ⎩ ⎧sin 4x ≠ 0 ⎪ ⇔⎨ π 5π ⎪ x = 6 + k2π ∨ 6 + k2π, k ∈ ⎩ π 5π ⇔ x = + k2π ∨ x = + k2π, k ∈ 6 6

Tröôøng hôïp 2

Neáu ⎨
Baø i 159

⎧A ≤ M ≤ B thì A = B = M ⎩A = B

Phöông phaùp ñoái laäp

Giaû i phöông trình:

sin4 x − cos4 x = sin x + cos x (*)

Ta coù : (*) ⇔ sin2 x − cos2 x = sin x + cos x
⇔ − cos 2x = sin x + cos x ⎧cos 2x ≤ 0 ⎪ ⇔⎨ 2 ⎪cos 2x = 1 + 2 sin x cos x ⎩ ⎧cos 2x ≤ 0 ⎧cos 2x ≤ 0 ⎪ ⇔⎨ ⇔⎨ 2 ⎪− sin 2x = 2 sin 2x ⎩sin 2x = 0 (cos 2x = ± 1 ) ⎩ ⇔ cos 2x = −1 ⇔x= π + kπ, k ∈ 2

Caù c h khaù c Ta coù sin 4 x − cos4 x ≤ sin4 x ≤ sin x ≤ sin x + cos x Do ñoù
⎧cos x = 0 π ⎪ ⇔ cos x = 0 ⇔ x = + kπ, k ∈ (*) ⇔ ⎨ 4 2 ⎪sin x = sin x ⎩

Baø i 160:

Giaû i phöông trình: ( cos 2x − cos 4x ) = 6 + 2 sin 3x (*)

2

Ta coù : (*) ⇔ 4 sin 2 3x.sin 2 x = 6 + 2 sin 3x • Do: sin 2 3x ≤ 1 vaø sin 2 x ≤ 1 neâ n 4 sin 2 3x sin 2 x ≤ 4 • Do sin 3x ≥ −1 neâ n 6 + 2 sin 3x ≥ 4 Vaä y 4 sin 2 3x sin 2 x ≤ 4 ≤ 6 + 2 sin 3x Daá u = cuû a phöông trình (*) ñuù n g khi vaø chæ khi

⎧sin2 3x = 1 ⎧sin2 x = 1 ⎪ 2 ⎨sin x = 1 ⇔ ⎨ ⎩sin 3x = −1 ⎪sin 3x = −1 ⎩ π ⎧ π ⎪ x = ± + k2π, k ∈ ⇔⎨ ⇔ x = + k2π, k ∈ 2 2 ⎪sin 3x = −1 ⎩
Baø i 161

cos3 x − sin 3 x = 2 cos 2x (*) sin x + cos x Ñieà u kieä n : sin x ≥ 0 ∧ cos x ≥ 0 Ta coù : (*) ⇔ ( cos x − sin x )(1 + sin x cos x ) = 2 ( cos2 x − sin 2 x ) sin x + cos x
Giaû i phöông trình:

(

)

⎡cos x − sin x = 0 ⇔⎢ ⎢1 + sin x cos x = 2 ( cos x + sin x ) sin x + cos x ⎣ π Ta coù : (1) ⇔ tgx = 1 ⇔ x = + kπ, k ∈ 4 Xeùt (2) Ta coù : khi sin x ≥ 0 thì sin x ≥ sin x ≥ sin 2 x Töông töï cos x ≥ cos x ≥ cos2 x sin x + cos x ≥ 1 vaø sin x + cos x ≥ 1 Vaä y Suy ra veá phaûi cuû a (2) thì ≥ 2 1 3 Maø veá traù i cuû a (2): 1 + sin 2x ≤ 2 2 Do ñoù (2) voâ nghieä m π Vaä y : (*) ⇔ x = + kπ, k ∈ 4

(

)

(1) (2)

Baø i 162:

Giaû i phöông trình:

3 − cos x − cos x + 1 = 2 (*)

Ta coù : (*)

⇔

3 − cos x = 2 + cos x + 1

⇔ 3 − cos x = 5 + cos x + 4 cos x + 1 ⇔ −2 ( cos x + 1) = 4 cos x + 1
Ta coù : −2 ( cos x + 1) ≤ 0 ∀x maø 4 cos x + 1 ≥ 0 ∀x Do ñoù daá u = cuû a (*) xaû y ra ⇔ cos x = −1 ⇔ x = π + k2π , k ∈

Baø i 163:

Giaû i phöông trình:

cos 3x + 2 − cos2 3x = 2 (1 + sin2 2x ) (*)

Do baá t ñaú n g thöù c Bunhiacoá p ski:

AX + BY ≤ A 2 + B2 . X 2 + Y 2
neâ n :
1 cos 3x + 1 2 − cos2 3x ≤ 2. cos2 3x + ( 2 − cos2 3x ) = 2

Daá u = xaû y ra ⇔ cos 3x = 2 − cos2 3x ⎧cos 3x ≥ 0 ⇔⎨ 2 2 ⎩cos 3x = 2 − cos 3x
⎧cos 3x ≥ 0 ⇔⎨ ⇔ cos 3x = 1 ⎩cos 3x = ±1 2 (1 + sin 2 2x ) ≥ 2

Maë t khaù c : Vaä y :

daá u = xaû y ra ⇔ sin 2x = 0

cos 3x + 2 − cos2 3x ≤ 2 ≤ 2 (1 + sin2 2x )

daá u = cuû a (*) chæ xaû y ra khi: cos 3x = 1 ∧ sin 2x = 0

⎧cos 3x = 1 ⎪ ⇔⎨ kπ ⎪ x = 2 , k ∈ ( coù 4 ñaàu ngoïn cung ) ⎩ ⇔ x = 2mπ , m ∈
Baø i 164: Giaû i phöông trình:
π⎞ ⎛ tg 2 x + cotg 2 x = 2 sin 5 ⎜ x + ⎟ (*) 4⎠ ⎝

Ñieà u kieä n : sin 2x ≠ 0 • Do baá t ñaú n g thöù c Cauchy: tg 2 x + cotg 2 x ≥ 2 daá u = xaû y ra khi tgx = cotgx π⎞ ⎛ • Maë t khaù c : sin ⎜ x + ⎟ ≤ 1 4⎠ ⎝ π⎞ ⎛ neâ n 2 sin5 ⎜ x + ⎟ ≤ 2 4⎠ ⎝ π⎞ ⎛ daá u = xaû y ra khi sin ⎜ x + ⎟ = 1 4⎠ ⎝ π⎞ ⎛ Do ñoù : tg 2 x + cotg 2 x ≥ 2 ≥ 2 sin5 ⎜ x + ⎟ 4⎠ ⎝ ⎧tgx = cotgx ⎪ Daá u = cuû a (*) xaû y ra ⇔ ⎨ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ = 1 ⎝ ⎠ ⎩

⎧tg 2 x = 1 ⎪ ⇔⎨ π ⎪ x = + k2π , k ∈ 4 ⎩ π ⇔ x = + k2π, k ∈ 4

Tröôøng hôïp 3: AÙp duïn g:

⎧ A ≤ M vaø B ≤ M ⎧A = M thì ⎨ ⎩A + B = M + N ⎩B = N ⎧sin u = 1 sin u + sin v = 2 ⇔ ⎨ ⎩sin v = 1 ⎧sin u = 1 sin u − sin v = 2 ⇔ ⎨ ⎩sin v = − 1 ⎧sin u = − 1 sin u + sin v = − 2 ⇔ ⎨ ⎩sin v = − 1

Neáu ⎨

Töông töï cho caù c tröôø n g hôïp sau

sin u ± cos v = ± 2 ; cos u ± cos v = ± 2

Baø i 165: Ta coù :

Giaû i phöông trình:

cos 2x + cos 3x =2 4

3x − 2 = 0 ( *) 4

( *) ⇔ cos 2x + cos

3x ≤1 4 neâ n daá u = cuû a (*) chæ xaû y ra ⎧ x = kπ , k ∈ ⎧cos 2x = 1 ⎪ ⎪ ⇔⎨ ⇔⎨ ⇔ x = 8mπ, m ∈ 8hπ 3x ⎪cos 4 = 1 ⎪x = 3 , h ∈ ⎩ ⎩ 8hπ 8h ⇔k= Do : kπ = 3 3 ñeå k nguyeân ta choïn h = 3m ( m ∈ Ζ ) ( thì k = 8m ) Do cos 2x ≤ 1 vaø cos

Caù c h khaù c ⎧cos 2x = 1 ⎪ ⇔ ⎨ 3x ⎪cos 4 = 1 ⎩ Baø i 166:

⎧ x = kπ , k ∈ ⎪ ⎨ 3kπ ⎪cos 4 = 1 ⎩

⇔ x = 8mπ, m ∈

Giaû i phöông trình: cos 2x + cos 4x + cos 6x = cos x.cos 2x.cos 3x + 2 ( *)

cos 2x + cos 4x + cos 6x

= 2 cos 3x cos x + 2 cos2 3x − 1 = 2 cos 3x ( cos x + cos 3x ) − 1

= 4 cos 3x.cos 2x.cos x − 1 1 Vaä y : cos 3x.cos 2x.cos x = ( cos 2x + 6 cos 4x + cos 6x + 1) 4 Do ñoù : 1 9 ( *) ⇔ cos 2x + cos 4x + cos 6x = ( cos2x + cos 4x + cos6x ) + 4 4 3 9 ⇔ ( cos 2x + cos 4x + cos 6x ) = 4 4 ⇔ cos 2x + cos 4x + cos 6x = 3 ⎧cos 2x = 1 ⎧2x = k2π, k ∈ (1) ⎪ ⎪ (2) ⇔ ⎨cos 4x = 1 ⇔ ⎨cos 4x = 1 ⎪cos 6x = 1 ⎪cos 6x = 1 (3) ⎩ ⎩ ⇔ 2x = k2π, k ∈ ⇔ x = kπ, k ∈ ( Theá (1) vaø o (2) vaø (3) ta thaá y hieå n nhieâ n thoû a ) Baø i 167: Giaû i phöông trình: cos 2x − 3 sin 2x − 3 sin x − cos x + 4 = 0 ( *)
Ta coù :

( *) ⇔ 2 = ⎜ − ⎜

⎞ ⎛ 1 ⎞ ⎛ 3 3 1 cos 2x + sin 2x ⎟ + ⎜ sin x + cos x ⎟ ⎟ ⎜ 2 ⎟ 2 2 ⎝ 2 ⎠ ⎝ ⎠

π⎞ π⎞ ⎛ ⎛ ⇔ 2 = sin ⎜ 2x − ⎟ + sin ⎜ x + ⎟ 6⎠ 6⎠ ⎝ ⎝ ⎧ π⎞ ⎛ π π ⎧ ⎪sin ⎜ 2x − 6 ⎟ = 1 ⎪2x − 6 = 2 + k2π, k ∈ ⎪ ⎝ ⎠ ⎪ ⇔⎨ ⇔⎨ ⎪sin ⎛ x + π ⎞ = 1 ⎪ x + π = π + h2π, h ∈ ⎜ ⎟ ⎪ ⎪ 6 2 ⎩ 6⎠ ⎝ ⎩

π ⎧ ⎪ x = 3 + kπ, k ∈ π ⎪ ⇔⎨ ⇔ x = + hπ, h ∈ 3 ⎪ x = π + h2π, h ∈ ⎪ 3 ⎩ Caù c h khaù c ⎧ π⎞ ⎛ ⎧ π⎞ ⎛ ⎪sin ⎜ 2x − 6 ⎟ = 1 ⎪sin ⎜ 2x − 6 ⎟ = 1 ⎪ ⎝ ⎠ ⎪ ⎝ ⎠ ( *) ⇔ ⎨ ⇔⎨ ⎪sin ⎛ x + π ⎞ = 1 ⎪ x + π = π + h2π, h ∈ ⎜ ⎟ ⎪ ⎪ 6⎠ 6 2 ⎩ ⎝ ⎩

⎧ π⎞ ⎛ ⎪sin ⎜ 2x − 6 ⎟ = 1 ⎪ ⎝ ⎠ ⇔⎨ ⎪ x = π + h2π, h ∈ ⎪ 3 ⎩
Baø i 168:

⇔x=

π + hπ, h ∈ 3

Giaû i phöông trình:

4 cos x − 2 cos 2x − cos 4x = 1 ( *)

Ta coù : ( * ) ⇔ 4 cos x − 2 ( 2 cos2 x − 1 ) − (1 − 2 sin 2 2x ) = 1

⇔ 4cosx − 4 cos2 x + 8 sin2 x cos2 x = 0 ⇔ cos x = 0 hay 1 − cos x + 2 sin 2 x cos x = 0
⇔ cos x = 0 hay 1 + cos x ( 2 sin 2 x − 1) = 0 ⇔ cos x = 0 hay 1 − cos x cos 2x = 0 ( * *) 1 ⇔ cos x = 0 hay 1 − ( cos 3x + cos x ) = 0 2 ⇔ cos x = 0 ∨ cos 3x + cos x = 2 ⎧cos 3x = 1 ⇔ cos x = 0 ∨ ⎨ ⎩cos x = 1

⎧cos x = 1 ⇔ cos x = 0 ⇔ ⎨ 3 ⎩4 cos x − 3 cos x = 1 ⇔ cos x = 0 ∨ cos x = 1 ⇔x=

π + kπ ∨ x = k2π, k ∈ 2 Caù c h khaù c ( * *) ⇔ cos x = 0 hay cos x cos 2x = 1 ⎧cos x = 1 ⎧cos x = − 1 ⇔ cos x = 0 ∨ ⎨ ∨⎨ ⎩cos 2x = 1 ⎩cos 2x = − 1 ⎧ x = k2π, k ∈ ⎧ x = π + k2π, k ∈ π ⇔ x = + kπ, k ∈ ∨ ⎨ ∨⎨ 2 ⎩cos 2x = 1 ⎩cos 2x = − 1 π ⇔ x = + kπ ∨ x = k2π, k ∈ 2 Baø i 169: Giaû i phöông trình: 1 tg2x + tg3x + = 0 ( *) sin x cos 2x cos 3x

( loaïi )

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

⇔ sin x ( sin 2x cos 3x + sin 3x cos 2x ) + 1 = 0

⇔ sin x.sin 5x = −1 1 ⇔ − ( cos 6x − cos 4x ) = −1 2 ⇔ cos 6x − cos 4x = 2 ⎧cos 6x = 1 ⇔⎨ ⇔ ⎩cos 4x = −1
Do ñoù : (*) voâ nghieä m . Caù c h khaù c

⎧t = cos 2x ⎪ 3 ⎨4t − 3t = 1 ⇔ ⎪ 2 ⎩2t − 1 = −1

⎧t = cos 2x ⎪ 3 ⎨4t − 3t = 1 ⎪ ⎩t = 0

⎧sin x = 1 ⎧sin x = − 1 hay ⎨ ⇔ sin x. sin 5x = −1 ⇔ ⎨ ⎩sin 5x = − 1 ⎩sin 5x = 1 π π ⎧ ⎧ ⎪ x = + k2π, k ∈ ⎪ x = − + k2π, k ∈ hay ⎨ ⇔⎨ 2 2 ⎪sin 5x = − 1 ⎪sin 5x = 1 ⎩ ⎩
⇔ x ∈∅

Baø i 170:

Giaû i phöông trình: cos2 3x.cos 2x − cos2 x = 0 ( *)
1 1 (1 + cos 6x ) cos 2x − (1 + cos 2x ) = 0 2 2 ⇔ cos 6x cos 2x = 1

Ta coù : ( * ) ⇔

1 ( cos 8x + cos 4x ) = 1 2 ⇔ cos 8x + cos 4x = 2 ⇔ ⎧cos 8x = 1 ⇔⎨ ⎩cos 4x = 1 ⎧2 cos2 4x − 1 = 1 ⇔⎨ ⎩cos 4x = 1 ⎧cos2 4x = 1 ⇔⎨ ⎩cos 4x = 1 ⇔ cos 4x = 1 ⇔ 4x = k2π, k ∈ kπ ,k ∈ ⇔x= 2 Caù c h khaù c ⇔ cos 6x cos 2x = 1 ⎧cos 2x = 1 ⎧cos 2x = −1 ⇔⎨ hay ⎨ ⎩cos 6x = 1 ⎩cos 6x = −1

⎧2x = k2π, k ∈ ⎧2x = π + k2π, k ∈ ⇔⎨ hay ⎨ ⎩cos 6x = 1 ⎩cos 6x = −1 kπ x= ,k ∈ 2 Caù c h khaù c ⎧cos 8x = 1 ⎧cos 8x = 1 ⇔⎨ ⎨ ⎩cos 4x = 1 ⎩4x = k2π, k ∈ kπ ⇔x= ,k ∈ 2

Do ñoù ta coù
sin x cos x
m

Tröôøng hôïp 4: DUØNG KHAÛO SAÙT HAØM SOÁ x y = a laø haøm giaûm khi 0< a <1.
< sin x < co s x
n

⇔ n > m, ∀x ≠
n

π
2

+ kπ , k ∈ + kπ , k ∈

m

⇔ n > m, ∀x ≠

π
2

sin x cos x

m m

≤ sin x ≤ co s x

n

⇔ n ≥ m, ∀x
n

⇔ n ≥ m, ∀x

Baø i 171:

Giaû i phöông trình: 1 −

x2 = cos x ( *) 2

Ta coù : ( *) ⇔ 1 = Xeù t

x2 + cos x treân R 2 Ta coù : y ' = x − sin x vaø y '' = 1 − cos x ≥ 0 ∀x ∈ R Do ñoù y’(x) laø haø m ñoà n g bieá n treâ n R Vaä y ∀x ∈ ( 0, ∞ ) : x > 0 neân y ' ( x ) > y ' ( 0) = 0 y=

x2 + cos x 2

∀x ∈ ( −∞, 0) : x < 0 neân y ' ( x ) < y ' ( 0) = 0
Do ñoù :

Vaä y : y =

x2 + cos x ≥ 1 ∀x ∈ R 2 Daá u = cuû a (*) chæ xaû y ra taï i x = 0 Do ñoù ( *) ⇔ x = 0 •

Baø i 172: Giaû i phöông trình
sin 4 x + sin 6 x = sin 8 x + sin10 x (*)

Ta coù
⎧sin 4 x ≥ sin 8 x vaø daáu =xaûy ra khi vaø chæ khi sin 2 x = 1hay sinx = 0 ⎪ ⎨ 6 10 2 ⎪ sin x ≥ sin x vaø daáu =xaûy ra khi vaø chæ khi sin x = 1 hay sinx = 0 ⎩

⇔ sin 2 x = 1 ∨ sinx = 0 π ⇔ x = ± + k 2π ∨ x = k 2π , k ∈ 2

Caù c h khaù c (*) ⇔ sin 4 x = 0 hay 1+ sin 2 x = sin 4 x + sin 6 x
⇔ sin x = 0 hay sin 2 x =1

Giaû i caù c phöông trình sau
1. 2. 3.

BAØI TAÄP

lg ( sin2 x ) − 1 + sin 3 x = 0

4. 5. 6.
7. 8. 9. 10.
11. 12. 13. 14. 15. 16. 17. 18.

π⎞ ⎛ sin 4x − cos 4x = 1 + 4 2 sin ⎜ x − ⎟ 4⎠ ⎝ 1 sin 2 x + sin 2 3x = sin x. sin 2 3x 4 sin x π = cos x

2 cos x + 2 sin 10x = 3 2 + 2 cos 28x. sin x

( cos 4x − cos 2x )

2

= 5 + sin 3x

sin x + cos x = 2 ( 2 − sin 3x ) sin 3x ( cos 2x − 2 sin 3x ) + cos 3x (1 + sin 2x − 2 cos 3x ) = 0 tgx + tg2x = − sin 3x cos 2x 2 log a ( cot gx ) = log 2 ( cos x )
⎡ π⎤ 2sin x = cos x vôùi x ∈ ⎢0, ⎥ ⎣ 2⎦ 13 14 cos x + sin x = 1 cos 2x − cos 6x + 4 ( sin 2x + 1) = 0 sin x + cos x = 2 ( 2 − cos 3x ) sin3 x + cos3 x = 2 − sin4 x cos2 x − 4 cos x − 2x sin x + x 2 + 3 = 0 sin x 2 + sin x = sin 2 x + cos x 3 cot g 2 x + 4 cos2 x − 2 3 cot gx − 4 cos x + 2 = 0
Th.S Phạm Hồng Danh (TT luyện thi Vĩnh Viễn)

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