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1) sinx – cosx = 1 |: 2 ;sin x ⋅ππ2;− sin cos x =442ππx − = (− 1)k + πk ;44π2;sin( x − ) =42π πx = (− 1)k + + πk , k ∈ Z ;4 42) sinx + cosx = 1 |: 2 ;sin x ⋅ππ2;+ sin cos x =442ππx + = (− 1)k + πk ;44π2;sin( x + ) =42π πx = (− 1)k − + πk , k ∈ Z ;4 4sin x cossin x cos3)31sin x + cos x = 1 ;22πsin( x + ) = 1 ;6πx = + 2πk , k ∈ Z ;33 sin x + cos x = 2 |:2;ππsin x + sin cos x = 1 ;66π πx + = + 2πk ;6 2cos4) sin 3x + cos 3x = 2 |: 2 ;ππsin 3x + cos 3x sin = 1 ;44π π3x + = + 2πk ;4 2cos222;+cos x =22222sin 3x +cos 3x = 1 ;22πsin(3 x + ) = 1 ;4π 2π+x=k, k ∈ Z .12 3626.

1) cosx = cos3x;cos3x – cosx = 0; – 2sin2xsinx = 0; sin2x = 0 илиπsinx = 0; 2x = πk или x = πk, k ∈ Z;x = k или x = πk (входит в серию2ππкорней x = k ), k ∈ Z,т.е. x = k , k ∈ Z ;222) sin5x = sinx; sin5x – sinx = 0; 2sin2xcos3x = 0; sin2x = 0 или cos3x = 0;πππ π2x = πk или 3x = + πk , k ∈ Z ;x = k или x = + k , k ∈ Z ;226 3π3).

sin 2x = cos3x; cos3x − sin 2x = 0; sin( + 3x) − sin 2x = 0 ;2169www.5balls.ruπ x π 5x  π x   π 5x sin  +  = 0 или cos +2 sin  +  cos +=0;=0;4 24 2 4 4 4 2 ππ 2ππ xπ 5x π+= + πk , k ∈ z ; x = − + 2πk или x =k, k ∈ z ;+ = πk или +4 24 22210 54). sin x + cos3x = 0; cos3x + cos( π − x) = 0 ;2πππ2cos( + x) cos( − + 2x) = 0; cos( + x) = 0 или444ππππ πcos(2x − ) = 0;+ x = + πk или 2 x − = + πk ,4424 2π3π πk ∈ z; x = + πk или x =+ k, k ∈ z .42 2627.

1) cos3x – cos5x = sin4x; – 2sin4xsin( – x) = sin4x; sin4x(1–2sinx)=0;1πsin4x = 0 или sin x = ; 4x = πk или x = (− 1)k + πk , k ∈ Z ;62ππx = k или x = (− 1)k + πk , k ∈ Z ;462) sin7x – sinx = cos4x; 2sin3xcos4x = cos4x; cos4x(2sin3x – 1) = 0;1ππ4 x = + πk или 3x = (− 1)k + πk , k ∈ Z ;cos4x = 0 или sin 3x = ;262π πππx = + k или x = (− 1)k+ k, k ∈ Z ;18 34 23) cosx + cos3x = 4cos2x; 2cos2xcos( – x) = 4cos2x; cos2x(4 – 2cosx) = 0;cos2x = 0 или cosx = 2;2x =π+ πk , k ∈ Z , во втором случае реше2ний нет, т.е.

x = π + π k , k ∈ Z ;424) sin2x – cos2x = cos4x; – cos2x = 2cos22x – 1; 2cos22x + cos2x – 1 = 0;11cos2x = a; 2a2 + a – 1 = 0; a1 = – 1, a 2 = ; cos2x = – 1 или cos 2 x = ;22π2x = π + 2πk или 2x = ± π + 2πk , k ∈ Z ; x = + πk или x = ± π + πk , k ∈ Z .263628. 1) (tgx − 3)(2sin x + 1) = 0 ;12tgx = 3 или sin x = − 1 ;122ππx= (− 1)k +1 + πk , k ∈ Z ;x = + πk или3126k +1πx = + πk или x = (− 1) 2π + 12πk , k ∈ Z ;32) (1 − 2 cos x )(1 + 3tgx) = 0 ;4cosx2или tgx = − 3 ;=423170www.5balls.ruxππ= ± + 2πk или x = − + πk , k ∈ Z ;446πх = ±π + 8πk, k ∈ Z или x = − + πk , k ∈ Z ;63) (2sin( x + π ) − 1)(2tgx + 1) = 0 ;1π1sin(x + ) = или tgx = − ;2626x+1ππ= (− 1)k + πk или x = – arctg + πk, k ∈ Z ;662x = (− 1)k1π π− + πk или x = – arctg + πk, k ∈ Z ;6 62π2или tgx = 3;cos(x + ) = −424) (1 + 2 cos(x + π ))(tgx − 3) = 0 ;4π3π=±+ 2πk или x = arctg3 + πk, k ∈ Z44πx = + 2πk , x = – π + 2πk или x = arctg3 + πk, k ∈ Z2x+первая серия корней не подходит, т.к.

tg( π + 2πk) — не существует, т.е.2x = – π + 2πk или x = arctg3 + πk, k ∈ Z629. 1)3 sin x cos x = sin 2 x ;3 − tgx = 0 ;πx = πk или x = + πk , k ∈ Z ;sinx = 0 или tgx = 3 ;32) 2sinxcosx = cosx;cosx(2sinx – 1) = 0;π1πx = + πk или x = (− 1)k + πk , k ∈ Z ;cosx = 0 или sinx = ;6222sin2xcos2x + sin22x = 0;3) sin4x + sin22x = 0;sin2x(2cos2x + sin2x) = 0;sin2x = 0 или 2cos2x + sin2x = 0;sin2x = 0 или 2 + tgx = 0;sin2x = 0 или tg2x = – 2;2x = πk или 2x = – arctg2 + πk, k ∈ Z;π1πx = k или x = − arctg2 + k , k ∈ Z ;2222sinxcosx + 2cos2x = 0;4) sin2x + 2cos2x = 0;2cosx(sinx + cosx) = 0;cosx = 0 или sinx + cosx = 0;cosx = 0 или tgx + 1 = 0;cosx = 0 или tgx = – 1;ππx = + πk или x = − + πk , k ∈ Z .24122630.

1) 2 sin x = 1 + sin 4x ;1 − cos 2 x = 1 + sin 2 x cos 2x ;33sinx = 0 или3 cos x − sin x = 0 ;sin x( 3 cos x − sin x) = 0 ;sinx = 0 или171www.5balls.ru23cos 2 x( sin 2 x + 1) = 0 ;cos2x = 0 или sin 2 x = − ;23ππ π2 x = + πk , во втором случае решений нет x = + k , k ∈ Z ;24 22) 2cos22x – 1 = sin4x;1 + cos4x – 1 = sin4x |:cos4x;ππ π1 = tg4x;4 x = + πk ;x=+ k, k ∈ Z ;416 433) 2cos22x + 3cos2x = 2;2 cos 2 x + (1 + cos 2 x ) = 2 ;2cos2x = a;4cos22x + 3cos2x – 1 = 0;11cos2x = – 1 или cos 2 x = ;4a2 + 3a – 1 = 0;a1 = – 1, a 2 = ;4412x = π + 2πk или 2 x = ± arccos + 2πk , k ∈ Z , т.е.4π11x = + πk или x = ± arccos + πk , k ∈ Z ;2244) (sinx + cosx)2 = 1 + cosx;sin2x + cos2x + 2sinxcosx = 1 + cosx;12sinxcosx = cosx;cosx(2sinx – 1) = 0; cosx = 0 или sin x = ;2πk πx = + πk или x = (− 1)+ πk , k ∈ Z .26631. 1) 2sin2x – 3(sinx + cosx) + 2 = 0;2sin2x – 3(sinx + cosx) + 2(sin2x + cos2x) = 0;2sin2x – 3(sinx + cosx) + 2(sinx + cosx)2 – 2sin2x = 0;(sinx + cosx)(2sinx + 2cosx – 3) = 0;3sinx + cosx = 0 или sin x + cos x = ; tgx + 1 = 0 или sin( x + π ) = 3 ;242 2πtgx = – 1, во втором случае решений нет x = − + πk , k ∈ Z .42) sin2x + 3 = 3sinx + 3cosx;sin2x + cos2x + 2sinxcosx + 2 = 3(sinx + cosx);(sinx + cosx)2 + 2 = 3(sinx + cosx);a = 1, a = 2;sinx + cosx = a;a2 – 3a + 2 = 0;cosx + sinx = 1 или cosx + sinx = 2;π2или sin( x + π ) = 2 ;sin( x + ) =442x+во втором случае решений нет, т.е.

x = (− 1)ππ= (−1) k + πk , k ∈ Z ;44π π− + πk , k ∈ Z .4 43) sin2x + 4(sinx + cosx) + 4 = 0;sin2x + cos2x + 2sinxcosx + 4(sinx + cosx) + 3 = 0;(sinx + cosx)2 + 4(sinx + cosx) + 3 = 0;172www.5balls.rusinx + cosx = a;a2 + 4a + 3 = 0;sinx + cosx = – 1 или sinx + cosx = – 3;π1sin( x + ) = −42a = – 1, a = – 3;или sin(x + π ) = − 3 ; x + π = (− 1)k +1 π + πk , k ∈ Z , а во42втором случае решений нет, т.е. x = (− 1)k +144π π− + πk , k ∈ Z .4 44) sin2x + 5(cosx + sinx + 1) = 0;sin2x + cos2x + sinxcosx + 5(sinx + cosx) + 4 = 0;(sinx + cosx)2 + 5(sinx + cosx) + 4 = 0;sinx + cosx = a;a2 + 5a + 4 = 0;a1 = – 1, a2 = – 4;sinx + cosx = – 1 или sinx + cosx = – 4;π2или sin(x + π ) = −2 2 ; x + π = (− 1)k +1 π + πk , k ∈ Z , а воsin(x + ) = −44424π πk1+втором случае решений нет, т.е. x = (− 1)− + πk , k ∈ Z .4 4632.

1) 1 − cos(π − x ) + sin  π + x  = 0 ;2 21 + cosx + cosx = 0;2)cos x = −12π22 cos(x − ) = (sin x + cos x ) ;4x=±2((cosx + sinx)(1 – (sinx + cosx)) = 0;2π+ 2 πk, k ∈ Z ;322cos x +sin x) = (sin x + cos x) 2 ;;22sinx + cosx = 0 или sinx + cosx = 1;tgx + 1 = 0 или sin( x + π ) = 1 ; tgx = – 1 или x + π = (− 1)k π + πk, k ∈ Z ;4442πππx = − + πk или x = (− 1)k − + πk , k ∈ Z .44 4633. 1) 8sinxcosxcos2x = 1;2sin4x = 1;1sin 4 x = ;24sin2xcos2x = 1;4 x = (− 1)kπ+ πk ;6x = (− 1)kπ π+ k, k ∈ Z ;24 4(1 – sin4x) + cos2x = 0;2) 1 + cos2x = sin4x;222cos2x(1 + sin2x) + cos2x = 0;(1 – sin x)(1 + sin x) + cos x = 0;πcos2x(2 + sin2x) = 0;cosx = 0;x = + πk , k ∈ Z .2634. 1) 2cos2x + 3sin4x + 4sin22x = 0 |:cos22x;4tg22x + 6tg2x + 2 = 0; tg2x = a;2a2 + 3a + 1 = 0;a1 = – 1, a 2 = −1;2π11; 2 x = − + πk или 2 x = −arctg + πk , k ∈ Z ;422π π11 πx = − + k или x = − arctg + k , k ∈ Z ;8 222 22) 1 – sinxcosx + 2cos2x = 0;sin2x – sinxcosx + 3cos2x = 0 |:cos2x;tg2x – tgx + 3 = 0tgx = a;tg2x = – 1 или tg 2x = −173www.5balls.rua2 – a + 3 = 0;D < 0 — решений нет3) 2 sin 2 x + 1 cos3 2 x = 1 ;412cos 2x( cos x − 1) = 0 ;41 − cos 2 x +1cos3 2 x = 1 ;4cos2x = 0 или cos2x = 4;во втором случае решений нет, т.е.x=2x =π+ πk , k ∈ Z ,4аπ π+ k, k ∈ Z ;4 24) sin22x + cos23x = 1 + 4sinx;sin22x – sin23x = 4sinx;(sin2x – sin3x)(sin3x + sin2x) = 4sinx;− 2 sinx5x5xxxxxx5x5x⋅ 2 sincoscos = 8 sin cos2sin cos (4 + 2cos sin ) = 0 ;2222222222sin(4 + sin5x) = 0sinx = 0 или sin5x = – 4;x = πk, k ∈ Z, а второе уравнение решений не имеет, т.е.

x = πk, k ∈ Z.635. 1) cosxcos2x = sinxsin2x;cosxcos2x = 2sin2xcosx;2cosx(1 – 4sin2x) = 0;cosx(cos2x – 2sin x) = 0;1ππcosx = 0 или sin x = ± ;x = + πk или x = ± + πk , k ∈ Z ;2622) sin2xcosx = cos2xsinx;2cos2xsinx = cos2xsinx;sinx(cos2x – 2cos2x) = 0;2x = πk, k ∈ Z;sinx = 0, т.к. cos2x – 2cos x = 1, т.е.3) sin3x = sin2xcosx;sin2xcosx + cos2xsinx = sin2xcosx;sinxcos2x = 0;sinx = 0 или cos2x = 0;ππ πx = πk или 2 x = + πk , k ∈ Z , т.е.x = πk или x = + k , k ∈ Z ;24 24) cos5xcosx = cos4x;cos5xcosx = cos5xcosx + sin5xsinx;sin5xsinx = 0;sin5x = 0 или sinx = 05x = πk или x = πk, k ∈ Z;πx = πk или x = k , k ∈ Z (первая серия корней входит во вторую), т.е.5πx = k, k ∈ Z .5636.

1) 4sin2x – 5sinxcosx – 6cos2x = 0 |:cos2x;3a1 = − , a2 = 2;4tg2x – 5tgx – 6 = 0; tgx = a;4a2 – 5a – 6 = 0;433x = −arctg + πk или x = arctg2 + πk, k ∈ Z;tgx = − или tgx = 2;442) 3sin2x – 7sinxcosx + 2cos2x = 0 |:cos2x;13tg2x – 7tgx + 2 = 0; tgx = a;3a2 – 7a + 2 = 0; a1 = , a2 = 2;311x = arctg + πk или x = arctg2 + πk, k ∈ Z;tgx = или tgx = 2;333) 1 – 4sinxcosx + 4cos2x = 0;sin2x – 4sinxcosx + 5cos2x = 0 |:cos2x;tg2x – 4tgx + 5 = 0;174www.5balls.rutgx = a;a2 – 4a + 5 = 0;D < 0 — решений нет;4) 1 + sin2x = 2sinxcosx;2tg2x – 2tgx + 1 = 0;2sin2x – 2sinxcosx + cos2x = 0 |:cos2x;2tgx = a;2a – 2a + 1 = 0D < 0 — решений нет.637. 1) 4sin3x + sin5x – 2sinxcos2x = 0;4sin3x + sin5x + sinx – sin3x = 0;3sin3x + 2sin3xcos2x = 0;sin3x(3 + 2cos2x) = 0;sin3x = 0 или cos 2x = − 3 ;23x = πk, k ∈ Z, во втором случае решений нет, т.е.

x = π k , k ∈ Z ;32) 6cos2xsinx + 7sin2x = 0;6cos2xsinx + 14sinxcosx = 0;2sinx(3cos2x + 7cosx) = 0;sinx = 0 или 6cos2x + 7cosx – 3 = 0;cosx = a;312a1 = − ,a 2 = ;sinx = 0 или 6a + 7a – 3 = 0;2331sinx = 0 или cos 2 x = − или cos 2 x = ;231x = πk или 2 x = ± arccos + 2πk , k ∈ Z , а во втором случае решений нет,311т.е. x = πk или x = ± arccos + 2πk , k ∈ Z .23638. 1) sin2x + sin22x = sin23x;(sinx – sin3x)(sinx + sin3x) + sin2x ⋅ 2sinxcosx = 0;– 2sinxcos2x ⋅ 2sin2xcosx + sin2x ⋅ 2sinx ⋅ cosx = 0;2sinx ⋅ cosx ⋅ sin2x(1 – 2cos2x) = 0; sin22x(1 – 2cos2x) = 0;π12x = πk или 2 x = ± + 2πk , k ∈ Z ;sin2x = 0 или cos 2x = ;32ππx = k или x = ± + πk , k ∈ Z ;262) sinx(1 – cosx)2 + cosx(1 – sinx)2 = 2;sinx + cosx + sinxcosx(sinx + cosx) – 4sinxcosx = 2;(sin x + cos x) 2 − 1⋅ (sin x + cos x) = 2(sin x + cos x) 2 ;2tt 2sinx + cosx = t;(2 + (t 2 − 1) − 4t) = 0 ;(t − 4t + 1) = 0 ;22(sin x + cos x) +t1 = 0 или t 2 = 2 + 3 или t 3 = 2 − 3 ;sinx + cosx = 0 или sin x + cos x = 2 + 3 или sin x + cos x = 2 − 3 ;tgx = – 1 или sin(x + π ) = 2 + 3 или sin(x + π ) = 2 − 3 ;4x=−242π2− 3π+ πk или x = − + (− 1)k arcsin+ πk , k ∈ Z , ;442175www.5balls.ruа во втором случае решений нет.639.

1) sin x sin 2x sin 3x = 1 sin 4 x ;4sinxsin2xsin3x = sinxcosxcos2x;sinx(cosxcos2x – sin2xsin3x) = 0;111111sin x( cos3x + cos x + cos5x − cos x) = 0 ; sin x( cos3x + cos5x) = 0 ;222222sinxcosxcos4x = 0;sinx = 0 или cosx = 0 или cos4x = 0;x = πk или x = π + πk или 4 x = π + πk , k ∈ Z ;22ππ πx = πk или x = + πk или x = + k , k ∈ Z ;8 4212) sin 4 x + cos 4 x = sin 2 2x ; (cos2x – sin2x)2 + 2sin2xcos2x = 2sin2xcos2x;2π2cos x = 0; 2 x = + πk , k ∈ Z ; x = π + π k , k ∈ Z .4 22640.

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