Nội dung

CHÖÔNG VII PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N VAØ PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A GIAÙ TRÒ TUYEÄT ÑOÁI A) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N Caù c h giaû i : AÙ p duï n g caù c coâ n g thöù c ⎧A ≥ 0 A = B⇔⎨ ⇔ ⎩A = B ⎧B ≥ 0 ⎨ ⎩A = B ⎧B ≥ 0 A =B⇔⎨ 2 ⎩A = B Ghi chuù : Do theo phöông trình chænh lyù ñaõ boû phaà n baá t phöông trình löôï n g giaù c neâ n ta xöû lyù ñieà u kieä n B ≥ 0 baè n g phöông phaù p thöû laï i vaø chuù n g toâ i boû caù c baø i toaù n quaù phöù c taï p . Baø i 138 : Giaû i phöông trình ( *) ⇔ 5 cos x − cos 2x + 2 sin x = 0 ( *) 5 cos x − cos 2x = −2 sin x ⎧sin x ≤ 0 ⇔⎨ 2 ⎩5 cos x − cos 2x = 4 sin x ⎧⎪sin x ≤ 0 ⇔⎨ 2 2 ⎪⎩5 cos x − 2 cos x − 1 = 4 1 − cos x ⎧sin x ≤ 0 ⇔⎨ 2 ⎩2 cos x + 5 cos x − 3 = 0 ⎧sin x ≤ 0 ⎪ ⇔⎨ 1 ⎪⎩cos x = 2 ∨ cos x = −3 ( loaïi ) ⎧sin x ≤ 0 ⎪ ⇔⎨ π ⎪⎩ x = ± 3 + k2π, k ∈ π ⇔ x = − + k2π, k ∈ 3 ( ) ( ) Baø i 139 : Giaû i phöông trình sin3 x + cos3 x + sin3 x cot gx + cos3 xtgx = 2 sin 2x Ñieà u kieä n : ⎧cos x ≠ 0 ⎪ ⎨sin x ≠ 0 ⇔ ⎪sin 2x ≥ 0 ⎩ ⎧sin 2x ≠ 0 ⇔ sin 2x > 0 ⎨ ⎩sin 2x ≥ 0 Luù c ñoù : ( *) ⇔ sin3 x + cos3 x + sin2 x cos x + cos2 x sin x = 2 sin 2x ⇔ sin2 x ( sin x + cos x ) + cos2 x ( cos x + sin x ) = 2sin 2x ( ) ⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin 2x ⎧⎪sin x + cos x ≥ 0 ⇔⎨ 2 ⎪⎩( sin x + cos x ) = 2 sin 2x ⎧ π⎞ ⎛ ⎧ π⎞ ⎛ ⎪ 2 sin ⎜ x + ⎟ ≥ 0 ⎪sin ⎜ x + ⎟ ≥ 0 4⎠ ⇔⎨ ⇔⎨ 4⎠ ⎝ ⎝ ⎪1 + sin 2x = 2 sin 2x ⎪sin 2x = 1 ( nhaän do sin 2x > 0 ) ⎩ ⎩ ⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪ x = π + kπ, k ∈ ⎪ x = π + m2π ∨ x = 5π + m2π ( loaïi ) , m ∈ ⎩⎪ 4 ⎩⎪ 4 4 π ⇔ x = + m2π, m ∈ 4 Baø i 140 : Giaû i phöông trình π⎞ ⎛ 1 + 8 sin 2x. cos2 2x = 2 sin ⎜ 3x + ⎟ ( *) 4⎠ ⎝ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ Ta coù : (*) ⇔ ⎨ ⎪1 + 8 sin 2x cos2 2x = 4 sin2 ⎛ 3x + π ⎞ ⎜ ⎟ ⎪⎩ 4⎠ ⎝ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪1 + 4 sin 2x (1 + cos 4x ) = 2 ⎡1 − cos( 6x + π ) ⎤ ⎢⎣ ⎪⎩ 2 ⎥⎦ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⇔⎨ ⎝ ⎪1 + 4 sin 2x + 2 ( sin 6x − sin 2x ) = 2 (1 + sin 6x ) ⎩ ⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪sin 2x = 1 ⎪ x = π + kπ ∨ x = 5π + kπ, k ∈ ⎩⎪ 2 ⎩⎪ 12 12 π⎞ ⎛ So laï i vôù i ñieà u kieä n sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⎝ π •Khi x = + kπ thì 12 π⎞ ⎛ ⎛π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = cos kπ 4⎠ ⎝ ⎝2 ⎠ ⎡1 , ( neáu k chaün ) ( nhaän ) =⎢ ⎢⎣ −1 , ( neáu k leû ) ( loaïi ) 5π • Khi x = + kπ thì 12 π⎞ ⎛ ⎛ 3π ⎞ ⎛ π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = sin ⎜ − + kπ ⎟ 4⎠ ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎡ −1 , neáu k chaün ( loaïi ) =⎢ ⎢⎣1 , neáu k leû ( nhaän ) π 5π + m2π ∨ x = + ( 2m + 1) π, m ∈ Do ñoù ( *) ⇔ x = 12 12 Baø i 141 : Giaû i phöông trình 1 − sin 2x + 1 + sin 2x = 4 cos x ( * ) sin x Luù c ñoù : ( *) ⇔ 1 − sin 2x + 1 + sin 2x = 2 sin 2x ( hieå n nhieâ n sinx = 0 khoâ n g laø nghieä m , vì sinx =0 thì VT = 2, VP = 0 ) ⎪⎧2 + 2 1 − sin2 2x = 4 sin2 2x ⇔⎨ ⎪⎩sin 2x ≥ 0 ⎧⎪ 1 − sin2 2x = 2 sin2 2x − 1 ⇔⎨ ⎪⎩sin 2x ≥ 0 ⎧1 − sin 2 2x = 4 sin4 2x − 4 sin2 2x + 1 ⎪ 1 ⎪ ⇔ ⎨sin2 2x ≥ 2 ⎪ ⎪⎩sin 2x ≥ 0 ⎧sin 2 2x 4 sin 2 2x − 3 = 0 ⎪ ⇔⎨ 1 ⎪sin 2x ≥ 2 ⎩ ⎧ 3 − 3 ∨ sin 2x = ⎪sin 2x = ⎪ 2 2 ⇔⎨ ⎪sin 2x ≥ 2 ⎪⎩ 2 3 ⇔ sin 2x = 2 ( ) π 2π + k2π ∨ 2x = + k2π, k ∈ 3 3 π π ⇔ x = + kπ ∨ x = + kπ, k ∈ 6 3 Chuù yù : Coù theå ñöa veà phöông trình chöù a giaù trò tuyeä t ñoá i ⎧sin x ≠ 0 ( *) ⇔ ⎪⎨ ⎪⎩ cos x − sin x + cos x + sin x = 2 sin 2x ⇔ cos x − sin x + cos x + sin x = 2 sin 2x ⇔ 2x = Baø i 142 : Giaû i phöông trình sin x + 3 cos x + sin x + 3 cos x = 2 ( * ) π 3 cos x Ñaët t = sin x + 3 cos x = sin x + π cos 3 π⎞ π⎞ 1 ⎛ ⎛ ⇔t= sin ⎜ x + ⎟ = 2 sin ⎜ x + ⎟ π 3⎠ 3⎠ ⎝ ⎝ cos 3 ( *) thaønh t + t = 2 sin ⇔ t = 2−t ⎧2 − t ≥ 0 ⎧t ≤ 2 ⇔⎨ ⇔ ⎨ 2 2 ⎩t = 4 − 4t + t ⎩t − 5t + 4 = 0 ⎧t ≤ 2 ⇔⎨ ⇔ t =1 ⎩t = 1 ∨ t = 4 Do ñoù ( * ) π⎞ 1 π π π 5π ⎛ ⇔ sin ⎜ x + ⎟ = ⇔ x + = + k2π hay x + = + k2π, k ∈ 3⎠ 2 3 6 3 6 ⎝ π π ⇔ x = − + k2π ∨ x = + k2π, k ∈ 6 2 Baø i 143 : Giaû i phöông trình 3 tgx + 1 ( sin x + 2 cos x ) = 5 ( sin x + 3 cos x ) ( *) Chia hai veá cuû a (*) cho cos x ≠ 0 ta ñöôï c ( *) ⇔ 3 tgx + 1 ( tgx + 2) = 5 ( tgx + 3) Ñaët u = tgx + 1 vôùi u ≥ 0 Thì u 2 − 1 = tgx ( ) ( (*) thaøn h 3u u 2 + 1 = 5 u 2 + 2 ) ⇔ 3u 3 − 5u 2 + 3u − 10 = 0 ⇔ ( u − 2 ) ( 3u 2 + u + 5 ) = 0 ⇔ u = 2 ∨ 3u 2 + u + 5 = 0 ( voâ nghieäm ) Do ñoù ( *) ⇔ tgx + 1 = 2 ⇔ tgx + 1 = 4 π π⎞ ⎛ ⇔ tgx = 3 = tgα ⎜ vôùi − < α < ⎟ ⇔ x = α + k π , k ∈ 2 2⎠ ⎝ Baø i 144 : Giaû i phöông trình ( *) ⇔ ( ( ) 1 − cos x + cos x cos 2x = ) 1 sin 4x ( *) 2 1 − cos x + cos x cos 2x = sin 2x cos 2x ⎧cos x ≥ 0 ⇔⎨ hay ⎩cos 2x = 0 1 − cos x + cos x = sin 2x ⎧cos x ≥ 0 ⎪ hay ⎨sin 2x ≥ 0 ⎪ 2 ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ⎧cos x ≥ 0 ⎧cos x ≥ 0 ⎪ ⎪ hay ⎨sin 2x ≥ 0 ⇔⎨ π π ⎪⎩ x = 4 + k 2 , k ∈ ⎪ 2 ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ( VT ≥ 1 ≥ VP ) ⎧cos x ≥ 0 cos x ≥ 0 ⎪ ⎧ ⎪ ⎪sin 2x ≥ 0 ⇔⎨ hay ⎨ 2 π 5π ⎪⎩ x = ± 4 + hπ hay x = ± 4 + hπ, h ∈ ⎪sin 2x = 1 ⎩⎪(1 − cos x ) cos x = 0 π ⇔ x = ± + hπ, h ∈ 4 ⎧sin 2x = 1 ⎧sin 2x = 1 hay ⎨ hay ⎨ ⎩cos x = 0 ( ⇒ sin 2x = 0 ) ⎩cos x = 1 ( ⇒ sin x = 0 ⇒ sin 2x = 0 ) π ⇔ x = ± + hπ, h ∈ 4 ⎧cos x ≥ 0 ⎪ ⇔⎨ π ⎪⎩2x = 2 + kπ, k ∈ Baø i 145 : Giaû i phöông trình sin3 x (1 + cot gx ) + cos3 x (1 + tgx ) = 2 sin x cos x ( *) sin x + cos x ⎞ ⎛ cos x + sin x ⎞ 3 ⎟ + cos x ⎜ ⎟ = 2 sin x cos x sin x cos x ⎝ ⎠ ⎝ ⎠ ( *) ⇔ sin3 x ⎛⎜ ( ) ⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin x cos x ⎧sin x + cos x ≥ 0 ⇔⎨ ⎩1 + sin 2x = 2 sin 2x ⎧ π⎞ ⎛ sin ⎜ x + ⎟ ≥ 0 ⎪ ⎧sin x + cos x ≥ 0 ⎪ 4⎠ ⎝ ⇔⎨ ⇔⎨ ⎩sin 2x = 1 ⎪ x = π + kπ, k ∈ ⎪⎩ 4 ⎧ π⎞ ⎛ ⎪⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎝ ⎠ ⇔⎨ ⎪ x + π = π + kπ, k ∈ ⎪⎩ 4 2 ⎧ π⎞ ⎛ ⎪⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎝ ⎠ ⇔⎨ ⎪ x + π = π + h2π hay x + π = 3π + h2π, h ∈ ⎪⎩ 4 2 4 2 π ⇔ x = + h2π, h ∈ 4 Baø i 146 : Giaû i phöông trình cos 2x + 1 + sin 2x = 2 sin x + cos x ( *) π⎞ ⎛ Ñieà u kieä n cos 2x ≥ 0 vaø sin ⎜ x + ⎟ ≥ 0 4⎠ ⎝ Luù c ñoù : ( *) ⇔ ( cos x + sin x ) cos2 x − sin 2 x + 2 ⇔ cos2 x − sin 2 x + ( cos x + sin x ) + 2 cos 2x ⇔ cos x ( cos x + sin x ) + ( sin x + cos x ) cos 2x 2 = 2 cos x + sin x 2 ( cos x + sin x ) = 4 ( sin x + cos x ) = 2 ( sin x + cos x ) ⎡sin x + cos x = 0 ⇔⎢ ⎣cos x + cos 2x = 2 ⎡ tgx = −1 ⇔⎢ ⎢⎣ cos 2x = 2 − cos x ( * *) ⎡ tgx = −1 ⇔⎢ 2 ⎣cos 2x = 4 − 4 cos x + cos x ⇔ tgx = −1 ∨ cos2 x + 4 cos x − 5 = 0 ⇔ tgx = −1 ∨ cos x = 1 ∨ cos x = −5 ( loaïi ) π + kπ ∨ x = k2π, k ∈ 4 π ⎛ π⎞ Thöû laï i : • x = − + kπ thì cos 2x = cos ⎜ − ⎟ = 0 ( nhaän ) 4 ⎝ 2⎠ π⎞ ⎛ Vaø sin ⎜ x + ⎟ = sin kπ = 0 ( nhaän ) 4⎠ ⎝ • x = k2π thì cos 2x = 1 ( nhaän ) ⇔x=− π⎞ π ⎛ vaø cos ⎜ x + ⎟ = cos > 0 ( nhaän ) 4⎠ 4 ⎝ π Do ñoù (*) ⇔ x = − + kπ ∨ x = k2π, k ∈ 4 Chuù yù : Taï i (**) coù theå duø n g phöông trình löôï n g giaù c khoâ n g möï c ⎧cos x + cos 2x = 2 ⎪⎩sin x + cos x ≥ 0 ( * *) ⇔ ⎪⎨ ⎧cos x = 1 ⎪ ⇔ ⎨cos 2x = 2 cos2 x − 1 = 1 ⎪sin x + cos x ≥ 0 ⎩ ⎧cos x = 1 ⇔⎨ ⇔ x = 2kπ, k ∈ ⎩sin x + cos x ≥ 0 Caù c h khaù c ( *) ⇔ ⇔ cos2 x − sin 2 x + ( cos x + sin x ) (cos x + sin x).(cos x − sin x ) + 2 = 2 cos x + sin x ( cos x + sin x ) ⎧⎪cos x + sin x > 0 ⇔ cos x + sin x = 0 hay ⎨ ⎪⎩ cos x − sin x + ⎧⎪cos x + sin x > 0 ⇔ tgx = − 1 hay ⎨ ⎪⎩2 cos x + 2 cos 2x = 4 2 = 2 cos x + sin x ( cos x + sin x ) = 2 ⎧⎪cos x + sin x > 0 ⇔ tgx = − 1 hay ⎨ ⎪⎩cos x + cos 2x = 2 ⎧cos x = 1 π ⇔ x = − + kπ, k ∈ hay ⎨ 4 ⎩cos 2x = 1 π ⇔ x = − + kπ hay x = 2kπ, k ∈ 4 ( nhaä n xeù t : khi cosx =1 thì sinx = 0 vaø sinx + cosx = 1 > 0 ) 1. Giaû i phöông trình : a/ 1 + sin x + cos x = 0 4x − cos2 x cos 3 =0 b/ 1 − tg 2 x BAØI TAÄP c/ sin x + 3 cos x = 2 + cos 2x + 3 sin 2x sin 2 x − 2 sin x + 2 = 2 sin x − 1 3tgx e/ 2 3 sin x = − 3 2 sin x − 1 d/ sin2 2x + cos4 2x − 1 =0 f/ sin cos x g/ 8 cos 4x cos2 2x + 1 − cos 3x + 1 = 0 h/ sin x + sin x + sin2 x + cos x = 1 k/ 5 − 3sin 2 x − 4 cos x = 1 − 2 cos x l/ cos 2x = cos2 x 1 + tgx 2. Cho phöông trình : 1 + sin x + 1 − sin x = m cos x (1) a/ Giaû i phöông trình khi m = 2 b/ Giaû i vaø bieä n luaä n theo m phöông trình (1) 3. Cho f(x) = 3cos 6 2x + sin 42x + cos4x – m a/ Giaû i phöông trình f(x) = 0 khi m = 0 b/ Cho g ( x ) = 2 cos2 2x 3 cos2 2x + 1 . Tìm taá t caû caù c giaù trò m ñeå phöông trình f(x) = g(x) coù nghieä m . ( ÑS : 1 ≤ m ≤ 0 ) 4. Tìm m ñeå phöông trình sau coù nghieä m 1 + 2 cos x + 1 + 2sin x = m (ÑS : 1+ 3 ≤ m ≤ 2 1+ 2 ) B) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÙ C TRÒ TUYEÄ T ÑOÁ I Caù ch giaû i : 1/ Môû giaù trò tuyeä t ñoá i baè n g ñònh nghóa 2/ AÙ p duï n g • A = B ⇔ A = ±B ⎧B ≥ 0 ⎧B ≥ 0 ⎧A ≥ 0 ⎧A < 0 •A =B⇔⎨ ⇔⎨ 2 ⇔⎨ ∨⎨ 2 ⎩ A = ±B ⎩ A = B ⎩ A = −B ⎩A = B Baø i 147 : Giaû i phöông trình cos 3x = 1 − 3 sin 3x ( *) ⎧1 − 3 sin 3x ≥ 0 ( *) ⇔ ⎪⎨ 2 2 ⎪⎩cos 3x = 1 − 2 3 sin 3x + 3sin 3x 1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪1 − sin 2 3x = 1 − 2 3 sin 3x + 3 sin 2 3x ⎩ 1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪4 sin 2 3x − 2 3 sin 3x = 0 ⎩ 1 ⎧ ⎪sin 3x ≤ 3 ⎪ ⇔⎨ ⎪sin 3x = 0 ∨ sin 3x = 3 ⎪⎩ 2 ⇔ sin 3x = 0 ⇔x= kπ ,k ∈ 3 Baø i 148 : Giaû i phöông trình 3sin x + 2 cos x − 2 = 0 ( * ) ( *) ⇔ 2 cos x = 2 − 3sin x ⎧2 − 3sin x ≥ 0 ⇔⎨ 2 2 ⎩4 cos x = 4 − 12 sin x + 9 sin x 2 ⎧ ⎪sin x ≤ 3 ⇔⎨ ⎪4 1 − sin 2 x = 4 − 12 sin x + 9 sin 2 x ⎩ ( ⇔ ⇔ ⇔ ⇔ ) 2 ⎧ ⎪sin x ≤ 3 ⎨ ⎪13 sin2 x − 12 sin x = 0 ⎩ 2 ⎧ ⎪⎪sin x ≤ 3 ⎨ ⎪sin x = 0 ∨ sin x = 12 ⎪⎩ 13 sin x = 0 x = kπ, k ∈ Baø i 149 : Giaû i phöông trình sin x cos x + sin x + cos x = 1 ( * ) π⎞ ⎛ Ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝ Vôù i ñieà u kieä n : 0 ≤ t ≤ 2 Thì t 2 = 1 + 2sin x cos x t2 − 1 +t =1 Do ñoù (*) thaø n h : 2 ⇔ t 2 + 2t − 3 = 0 ⇔ t = 1 ∨ t = −3 ( loaïi ) Vaä y ( * ) ⇔ 12 = 1 + 2sin x cos x ⇔ sin 2x = 0 kπ ,k ∈ 2 Baø i 150 : Giaû i phöông trình ⇔x= ( sin x − cos x + 2 sin 2x = 1 ( * ) Ñaët t = sin x − cos x ñieàu kieän 0 ≤ t ≤ 2 Thì t = 1 − sin 2x ( *) thaønh : t + 2 1 − t 2 = 1 2 ( ) ⇔ 2t 2 − t − 1 = 0 1 ⇔ t = 1 ∨ t = − ( loaïi do ñieàu kieän ) 2 2 khi t = 1 thì 1 = 1 − sin 2x ) ⇔ sin 2x = 0 kπ ⇔x= ,k ∈ 2 Baø i 151 : Giaû i phuông trình sin 4 x − cos4 x = sin x + cos x ( * ) ( *) ⇔ ( sin2 x + cos2 x )( 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 ⇔⎨ 2 ⎪⎩1 − sin 2x = 1 + sin 2x ⎧⎪cos 2x ≤ 0 ⇔⎨ 2 ⎪⎩ sin 2x = − sin 2x ⎧cos 2x ≤ 0 ⇔⎨ ⎩sin 2x = 0 ⎧cos 2x ≤ 0 ⇔⎨ 2 ⇔ cos 2x = −1 ⎩cos 2x = 1 π ⇔ x = + kπ, k ∈ 2 Baø i 152 : Giaû i phöông trình 3 sin 2x − 2 cos2 x = 2 2 + 2 cos 2x ( *) ( Ta coù : ( * ) ⇔ 2 3 sin x cos x − 2 cos2 x = 2 2 + 2 2 cos2 x − 1 ) ⎛ 3 ⎞ 1 ⇔ cos x ⎜⎜ sin x − cos x ⎟⎟ = cos x 2 ⎝ 2 ⎠ π⎞ ⎛ ⇔ cos x.sin ⎜ x − ⎟ = cos x 6⎠ ⎝ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π⎞ π⎞ ⎛ ⎛ ⎪sin ⎜ x − 6 ⎟ = 1 ⎪sin ⎜ x − 6 ⎟ = −1 ⎝ ⎠ ⎝ ⎠ ⎩ ⎩ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π π π π ⎪⎩ x − 6 = 2 + k2π, k ∈ ⎪⎩ x − 6 = − 2 + k2π, k ∈ ⎧cos x > 0 ⎧cos x < 0 π ⎪ ⎪ ⇔ x = + kπ, k ∈ ∨ ⎨ ∨⎨ 2π π 2 ⎪⎩ x = 3 + k2π, k ∈ ⎪⎩ x = − 3 + k2π, k ∈ π ⇔ x = + kπ, k ∈ 2