Odpowiedzi

2010-03-21T19:28:08+01:00
A) tg (2x-π/8) = 1
2x-π/8=π/4 + kπ
2x=3π/8 + kπ
x=3π/16 + ½kπ

b) ctg (1/3x + π/3) = √3
1/3x + π/3 = π/6 + kπ
⅓x=-π/6 + kπ
x=-π/2 + 3kπ

c) 3 + 4cos (0,5x) = -1
4cos(0,5x)=-4
cos(0,5x)=-1
0,5x=π+2kπ
x=2π+4kπ

d) 2sin3x = -√2
sin3x = -√2/2
3x=-π/4 + 2kπ lub 3x=-3π/4 + 2kπ
x=-π/12 + ⅔kπ lub x=-π/4 + ⅔kπ

e) 3ctg (2x+π) = -√3
ctg(2x+π)=-√3/3
2x+π
2 2 2
Najlepsza Odpowiedź!
2010-03-21T19:30:07+01:00
A) tg (2x-π/8) = 1
2x-π/8=π/4 + kπ
2x=3π/8 + kπ
x=3π/16 + ½kπ

b) ctg (1/3x + π/3) = √3
1/3x + π/3 = π/6 + kπ
⅓x=-π/6 + kπ
x=-π/2 + 3kπ

c) 3 + 4cos (0,5x) = -1
4cos(0,5x)=-4
cos(0,5x)=-1
0,5x=π+2kπ
x=2π+4kπ

d) 2sin3x = -√2
sin3x = -√2/2
3x=-π/4 + 2kπ lub 3x=-3π/4 + 2kπ
x=-π/12 + ⅔kπ lub x=-π/4 + ⅔kπ

e) 3ctg (2x+π) = -√3
ctg(2x+π)=-√3/3
ctg(-2x-π)=√3/3
-2x-π=π/2
2x=-3π/2
x=-3π/4
5 3 5