tieỏt 10: Luyeọn taọp PHệễNG TRèNH LệễẽNG GIAC Cễ BAN TỐM TẮT Giả sử u, v là những biểu thức theo x. Ta có: + sin u = sin v ⇔ u = v + k2π u = π - v + k2π (k ∈ Z) + cos u = cos v ⇔ u = v + k2π u = - v + k2π (k ∈ Z) + tan u = tan v ⇔ u, v ≠ π/2 + mπ u = v + kπ (m, k ∈ Z) + cot u = cot v ⇔ u, v ≠ mπ u = v + kπ (m, k ∈ Z) !" π # $ # ! Gụùi yự traỷ lụứi a) sin (x - 3) = 1/4 x 3 = arcsin1/4 + k2 x - 3 = - arcsin1/4 + k2 x = 3 + arcsin1/4 + k2 x = 3 + - arcsin1/4 + k2 (k Z) b) sin 5x = 1 5x = /2 + k 2 (k Z) x = /10 + k 2/5 (k Z) !" π # ⇔ !" π % π % ∈ & ⇔ ! π "% π % ∈ & ⇔ π '"%! π % ∈ & ($ # ! ⇔ # # # ⇔ # "%!# # ⇔ ' # "%!# # % ∈ & BT2: Giaûi caùc phöông trình sau a) cos (x+2) = 2/5 b) cos 4x = cos 20 0 c) cos (2x/3 - π/3) = -1/2 d) tan (2x – 15 0 ) = 1; -180 0 ≤ x ≤ 120 0 )*+, -"!! ⇔ "!! "%! π "!! "%! π % ∈ & ⇔ !"! "%! π !.! "%! π % ∈ & ( • b) cos 4x = cos 20 0 • ⇔ 4x = 20 0 + k360 0 • 4x = -20 0 + k360 0 (k ∈ Z) • ⇔ x = 5 0 + k 90 0 • x = -5 0 + k 90 0 (k ∈ Z) c) cos (2x/3 - π/3) = -1/2 ⇔ cos (2x/3 - π/3) = cos (2π/3) ⇔ 2x/3 - π/3= 2π/3 + k 2π 2x/3 - π/3= - 2π/3 + k 2π (k ∈ Z) ⇔ 2x/3 = π + k 2π 2x/3 = - π/3+ k 2π (k ∈ Z) ⇔ x = 3π/2 + k 3π x = - π/2 + k 3π (k ∈ Z) $!. # # ⇔ !. # # "%/# # ⇔ # # "%0# # 1-2345%467- /# # ≤ # # "%0# # ≤ !# # ⇔ 8 ≤ % ≤ ⇒ % ∈ 9!77#7: ;6*46<+, # # 7'# # 7# # 7!# #