Goniometrické rovnice
(Seminář z matematiky i - m1130/02 2014)
(1) Řešte v IR rovnice:
(a) 2 sin * — sin* = 0
(b) sin2*+ 2sin*-3 =0
(c) cos*(2cos* + 1) = 1
(d) v/3tg2* + 2tg*- ^ = 0
(e) 3 sin* = 2cos2*
(f) sin* + cos2* = -
4
(g) 3cos2* — 4cos*— sin2* — 2 = 0
(h) cos 2* +sin* = 0
(i) sin* = cos*
(j) sin*+ \/3 cos* = 0
/i x sinjc
(k) —-pr + cos* = 1
(1) y/3sin* + cos* = v7^
* = U {f+ 2ta}
K= (J {^: + 2ta, f+2ta, |^: + 2ta}
^ = U {f +ta, f^ + ta}
£ = U {f +2ta, §rc + 2Jbr}
A:= U {g7T + 2ta, ^-^: + 2ta}
A:= U {f + 2ta, \% + 2k%, ^n + lkn}
K= U {f+ta}
* = U {|^+ta}
£ = U {2kK,§ + 2k7t}
ke2
K= (J {f +2ta,f + 2ta}
jfceZ
(2) Řešte v IR následující rovnice (nezapomeňte stanovit podmínky, je-li to třeba):
(a) sin2* + cos2* = 1 +tg*
(b) sin (*+f)+cos (*+§) = l+cos2*
(c) sin 2* — cos2 2* = cos 2*
K= U {ta,f+*f}
A:= U {f+ta, f+2ta, f^: + 2ta}
K= U {f+ta, |7r + ta, f+ta}
jfceZ
(d) COSJC +sinjc
cos2jc
1 — sin2jc (e) cos4x — sin4x = únlx sinjc + sin3jc
(f)
cosjc — cos3jc
(g) cosjccos2jc = cos4jccos5jc
(h) sin3jc = 2 siru
(i) tgx— sku + cosjc = 1
(j) 2sin2jc + cosjc = 2sin2jc COSJC+ 1
(k) ^±3^ = i
1 — V3tgx
3
(1) sin2 2x + sin2 4x= -
(m) |sku| = sinjc + 2 , 4
(n) |tgjc + cotgjc| = -j=
K= U {ln + kn,2kn, \n + 2kn}
K= U {f+*f}
jfceZ
K= U {1+kTt}
kel
k= u m
kel
kei
kel
* = U {f+*f,f+*f,f+*f}
K= (J {§rc + 2Jbr}
kel