Y=2 sinx-1
y=0 ⇒ 2 sinx -1=0
2 sinx=1
sinx= 1/2
x=(-1)ⁿ arcsin(1/2)+ πn; n ∈ Ζ
x= (-1)ⁿ × π/6; n ∈ Z
Ответ: x = (-1)ⁿ×π/6; n∈ Z
<span>17+12√2 / 2√2+3 -2√2 = 0
17+6+3-2</span>√2=0
26-2√2=0 - не может такого быть
3cos²2x - 5sin²x - sin2x = 0
5sin²x + sin2x - 3cos²x = 0
5sin²x + 2sinxcosx - 3cos²x = 0 |:cos²x
5tg²x + 2tgx - 3 = 0
5tg²x + 5tgx - 3tgx - 3 = 0
5tgx(tgx + 1) - 3(tgx + 1) = 0
(5tgx - 3)(tgx + 1) = 0
1) 5tgx - 3 = 0
5tgx = 3
tgx = 3/5
x = arctg(3/5) + πn, n ∈ Z
2) tgx + 1 = 0
tgx = -1
x = -π/4 + πk, k ∈ Z
Ответ: x = arctg(3/5) + πn, n ∈ Z; -π/4 + πk, k ∈ Z.