Luonggiac-Chuong8

# Luonggiac-Chuong8 - C HNG VIII P HNG TRNH L N G GIAC KHONG...

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CHÖÔNG VIII PHÖÔNG TRÌNH LÖÔÏNG GIAÙC KHOÂNG MAÃU MÖÏC Tröôøng hôïp 1 : TOÅNG HAI SOÁ KHOÂNG AÂM AÙp duïng Neáu A 0B0 AB0 ≥∧ ≥ += thì A = B = 0 Baøi 156 Giaûi phöông trình: 22 4cos x 3tg x 4 3cosx 2 3tgx 4 0 (*) +− + + = Ta coù: () ( ) ⇔− + + = =− π =± + π π ⇔= −+ π ∈ ± ± (*) 2 cos x 3 3tgx 1 0 3 cos x 2 1 tgx 3 xk 2 , k 6 1 tgx 3 2 , k 6 = Baøi 157 ( ) 2 8cos4x.cos 2x 1 cos3x 1 0 * + = () ( ) ++ + * 4cos4x 1 cos4x 1 1 cos3x 0 = ⇔+ + + + = ⎧⎧ ⎪⎪ ⇔⇔ ⎨⎨ == π ⎩⎩ ± 2 2 4cos 4x 0 2cos4x 1 0 11 cos 4x cos 4x cos 3x 1 3x k2 , k = π =∈ ± 1 cos 4x 2 k2 x , k (coù 3 ñaàu ngoïn cung) 3

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=− ππ π = π = + π π ⇔= ± + π ∈ ± ± 1 cos 4x 2 22 x + m 2 h a y x m 2 h a y xm 2 , m 33 2 2 , m 3 (ta nhaän = ± k1 vaø loaïi k = 0 ) Baøi 158 Giaûi phöông trình: () 2 23 3 sin 3x sin x cos3x sin x sin 3x cos x sin x sin 3x * 3sin4x ++ = 2 Ta coù: cos3x.sin 3x sin 3x.cos x + ( ) + + = == 3 3 2 4 c o sx 3 c o sxs i nx 3 s i nx 4 s i nxc o sx 3cos x sin x 3sin x cos x 3sin x cos x cos x sin x sin 2x.cos 2x sin 4x 24 2 ⇔+ = ⎛⎞ ⇔− + = ⎜⎟ ⎝⎠ + = 2 2 2 2 2 1 Vaäy: * sin x sin 3x sin x sin 3x vaø sin 4x 0 4 11 1 sin 3x sin x sin 3x sin 3x 0 vaø sin 4x 0 4 sin 3x sin x sin 3x 1 sin 3x 0 vaø sin 4x 0 += =∨ = 2 2 sin 3x sin x sin 6x 0 vaø sin 4x 0 21 6 sin 4x 0 1 sin 3x sin x 2 s i n 3 x0c o s 3 x0 ⎪⎪ = ⎨⎨ = = ± sin 4x 0 sin 4x 0 1 sin 3x 0 sin x 2 sin x 0 (VN) sin 3x 1 −= 3 sin 4x 0 1 sin x 2 3sinx 4sin x 1 ±
= ππ =+ π + π∈ = = ± ± sin 4x 0 1 sin x 2 sin 4x 0 5 xk 2 k 2 , k 66 5 2 x k 2 , k Tröôøng hôïp 2 Phöông phaùp ñoái laäp N e á u A MB AB ≤≤ = thì A BM = = Baøi 159 Giaûi phöông trình: −= + 44 sin x cos x sin x cos x (*) Ta coù: (*) ⇔−=+ 22 sin x cos x sin x cos x ⇔− = + ⇔⇔ ⎨⎨ = ⇔= π ⇔=+π∈ ± 2 2 cos 2x sin x cos x cos 2x 0 cos 2x 1 2 sin x cos x cos 2x 0 cos 2x 0 sin 2x 0 (cos 2x 1) sin 2x 2 sin 2x cos 2x 1 , k 2 Caùch khaùc Ta coù −≤ ≤≤+ 4 x cos x sin x sin x sin x cos x sin Do ñoù = = = 4 cos x 0 (*) cos x 0 sin x sin x π ± , k 2 Baøi 160: () 2 cos2x cos4x 6 2sin 3x (*) + 4 sin 3x.sin x 6 2sin 3x + Do: vaø 2 sin 3x 1 2 sin x 1 neân 4sin 3xsin x 4 Do 62 ≥− sin 3x 1 s i n 3 x4 + Vaäy 4 sin 3x sin x 4 6 Daáu = cuûa phöông trình (*) ñuùng khi vaø chæ khi

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= = =⇔ ⎨⎨ = − =− 2 2 2 sin 3x 1 sin x 1 sin x 1 sin 3x 1 sin 3x 1 π =± + π π ⇔⇔ = + ± π ± xk 2 , k 2 , k 2 2 sin 3x 1 Baøi 161 Giaûi phöông trình: 33 cos x sin x 2cos2x(*) sin x cos x = + Ñieàu kieän: si n x 0
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## This note was uploaded on 11/26/2011 for the course MATH 1002 taught by Professor Chuck during the Spring '11 term at University of Western States.

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Luonggiac-Chuong8 - C HNG VIII P HNG TRNH L N G GIAC KHONG...

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