6±5 D ⎛ x⎞ , = 36 − 11 = 25 , tg ⎜ ⎟ = 4 11 ⎝ 2 ⎠1, 2 π ⎡x π ⎡ ⎢ 2 = 4 + πn ⎢ x = 2 + 2πn ; k, n ∈ Z ⎢x ⎢ ⎢ = arctg 1 + πk ⎢ x = 2arctg 1 + 2πk ⎢⎣ 2 ⎢⎣ 11 11 π 1 Ответ: x = + 2πn; x = 2arctg + 2πk ; k, n ∈ Z . 11 2 ⎡ x ⎢tg 2 = 1 ⎢ x 1 ⎢tg = ⎢⎣ 2 11
№ 1187
1) tg3x + tg2x – 2tg x – 2 = 0, tg2x(tg x + 1) – 2(tg x + 1) = 0, π π ⎡ ⎡ ⎡tgx + 1 = 0 ⎢ x = − + nπ, n ∈ Z ⎢ x = − + nπ, n ∈ Z ; . 4 4 ⎢tg 2 x = 2 ; ⎢ ⎢ ⎣ ⎣⎢tgx = ± 2 ⎣⎢ x = ± arctg 2 + lπ, l ∈ Z Ответ: x = −
π + nπ, n ∈ Z , 4
x = ± arctg 2 + lπ, l ∈ Z .
sin x − sin x ; cos x ≠ 0, cos x 2 cos x – cos x = sin x – cos x ⋅ sin x, cos x(sin x – cos x) = sin x – cos x, ⎡ x = 2nπ, n ∈ Z ⎡cos = 1 ; ⎢ (cos x – 1)(sin x – cos x) = 0, ⎢ π ⎣sin x − cos x = 0 ⎢ x = + lπ, l ∈ Z 4 ⎣ π Ответ: x = 2nπ, n ∈ Z, x = + lπ, l ∈ Z . 4
2) 1 – cos x = tg x – sin x, 1 − cos x =
№ 1188
1) sin x + sin2x = cos x + 2cos2x, sin x(1 + 2cos x) = cos x(1 + 2cos x), (sin x – cos x)(1 + 2cos x) = 0, π ⎡ ⎢ x = 4 + nπ, n ∈ Z ⎢ ⎢ x = ±⎛⎜ π − π ⎞⎟ + 2lπ, l ∈ Z ⎢⎣ 3⎠ ⎝ π 2 Ответ: x = + nπ, n ∈ Z , x = ± π + 2lπ, l ∈ Z . 4 3
2) 2 cos2x = 6 (cos x − sin x) , 2(cos x − sin x )(cos x + sin x ) − 6 (cos x − sin x ) = 0 ,
(cos x − sin x )(2(cos x + sin x ) −
)
6 =0,
π ⎡ ⎡cos x − sin x = 0 ⎢ x = 4 + nπ, n ∈ Z ⎡cos x − sin x = 0 ⎢ ⎢ 2(cos x + sin x ) = 6 ⎢cos x + sin x = 3 ⎢ 3 ⎢ ⎣ ⎢⎣ 2 ⎢cos x + sin x = 2 ⎣
181