2cos2x+sinx–2)√5tgx=0
ОДЗ 5tgx > =0
(2cos2x+sinx–2)√5tgx=0
1ый корень √5tgx=0 = > x=πn
2cos2x+sinx–2 = 0
2(1–sin2x)+sinx–2 = 0
2–2sin2x+sinx–2 = 0
–2sin2x+sinx = 0
2sin2x–sinx = 0
sinx(2sinx–1) = 0
sinx = 0
2cos2x+sinx–2)√5tgx=0
ОДЗ 5tgx > =0
(2cos2x+sinx–2)√5tgx=0
1ый корень √5tgx=0 = > x=πn
2cos2x+sinx–2 = 0
2(1–sin2x)+sinx–2 = 0
2–2sin2x+sinx–2 = 0
–2sin2x+sinx = 0
2sin2x–sinx = 0
sinx(2sinx–1) = 0
sinx = 0