Задача 67563 Сам не могу решить ...
Условие
Решение
2 - 2cos^2 x + 3cos x - 3 = 0
2cos^2 x - 3cos x + 1 = 0
Получили квадратное уравнение относительно cos x:
(cos x - 1)(2cos x - 1) = 0
cos x = 1; [b]x1 = 2πn; n ∈ Z[/b]
cos x = 1/2; [b]x2 = +-π/3 + 2πk; k ∈ Z[/b]
2) 3sin^2 x + cos^2 x - 2 = 0
3sin^2 x + cos^2 x - 2sin^2 x - 2cos^2 x = 0
sin^2 x - cos^2 x = 0
sin^2 x = cos^2 x
tg^2 x = 1
tg x = -1; [b]x1 = -π/4 + πk; k ∈ Z[/b]
tg x = 1; [b]x2 = π/4 + πk; k ∈ Z[/b]
3) 7sin^2 x - 5cos^2 x + 2 = 0
7sin^2 x - 5cos^2 x + 2sin^2 x + 2cos^2 x = 0
9sin^2 x - 3cos^2 x = 0
3sin^2 x = cos^2 x
tg^2 x = 1/3
tg x = -1/sqrt(3); [b]x1 = -π/6 + πk; k ∈ Z[/b]
tg x = 1/sqrt(3); [b]x2 = π/6 + πk; k ∈ Z[/b]
4) 5ctg^2 x - 8ctg x + 3 = 0
Это квадратное уравнение относительно ctg x
D = 8^2 - 4*5*3 = 64 - 60 = 4 = 2^2
ctg x = (8 - 2)/10 = 6/10 = 0,6; [b]x1 = arcctg(0,6) + πk; k ∈ Z[/b]
ctg x = (8 + 2)/10 = 10/10 = 1; [b]x2 = π/4 + πk; k ∈ Z[/b]
5) tg x*(sin x + cos x) = 0
tg x = 0; [b]x1 = πk; k ∈ Z[/b]
sin x + cos x = 0
sin x = -cos x
tg x = -1; [b]x2 = -π/4 + πk; k ∈ Z[/b]