г) 2sin2x – sinxcosx = cos2x; 2tg2x – tgx – 1 = 0; tgx = t; 2t2 – t – 1 = 0; t = 1, t = –
1 ; 2
π + πk, k ∈ Z; 4 1 1 2) tgx = – , x = arctg − + πn, n ∈ Z; 2 2
1) tgx = 1, x =
π 4
1 2
Ответ: + πk / k ∈ Z ; arctg − + πn / n ∈ Z .
170. а) 4sin2x – sin2x = 3; sin2x – 2sinxcosx – 3cos2x=0; tg2x – 2tgx – 3 = 0; 1) tgx = –1, x = –
π + πn, n ∈ Z; 4
2) tgx = –3, x = arctg3+ πk, k ∈ Z; π + πn / n ∈ Z ; arctg3 + πk / k ∈ Z . 4
Ответ: −
б) cos2x = 2cosx – 1; 1 + cos2x – 2cosx = 0; cosx(cosx – 1) = 0; cosx = 0 или cosx = 1; 1) cosx = 0, x =
π + πk, k ∈ Z; 2
2) cosx = 1, x = 2πn, n ∈ Z. π 2
Ответ: + πk / k ∈ Z ; 2πn / n ∈ Z .
в) sin2x – cosx = 0; 2sinxcosx – cosx = 0; 1 ) = 0; 2 1 cosx = 0 или sinx = ; 2 π 1) cosx = 0, x = + πn, n ∈ Z; 2 1 π 2) sinx = , x = (–1)k + πk, k ∈ Z. 2 6 π Ответ: + πn / n ∈ Z ; 2πn / n ∈ Z . 2
2cosx(sinx –
93