Goniometrické rovnice
(Seminář z matematiky i - m1130/02 2017)
(1) Řešte v IR rovnice:
(a) 2 sin x — sin* = 0
(b) sin2*+ 2sin*-3 = 0
(c) cos*(2cos* + 1) = 1
(d) v/3tg2* + 2tg*-v/3 =0
(e) 3 sin* = 2cos2*
(f) sin* + cos2* = -
4
(g) 3cos2* — 4cos* — sin2* — 2 = 0
(h) cos2* + sin* = 0
(i) sin* = cos*
(j) sin*+ y/3 cos* = 0 /i x sin-*
(k) —=■ + cos* = 1 V3
(1) \/3sin* + cos* = y/3
K= U {ta, f + 2for, |^ + 2^}
£eZ
U {f +2^}
£eZ
fceZ
* = U {f + ta, jn + kn}
keZ
K= U {f + 2ta, |^: + 2^}
fceZ
keZ
keZ
K= U {%+2kn,ln + 2kn,^n + 2kn}
keZ
K= U {f+ ^}
£eZ
* = U {i^ + ta}
£eZ
U {2ta,§ + 2kx)
keZ
K= U {f + 2for,f + 2ta}
£eZ
(2) Řešte v IR následující rovnice (nezapomeňte stanovit podmínky, je-li to třeba):
(a) sin2* + cos2* = 1 +tg*
(b) sin(*+|)+cos(*+|) = l+cos2*
(c) sin 2* — cos2 2* = cos 2*
K= U {ta,f+ fcf}
fceZ
keZ
K= U {%+k7C,%7C + k7C,%+k7c}
keZ
(d) cosA' + sinx
cos2a"
1 — sin2A" (e) cos4x — sin4x = ún2x sinx+ sin3jc
(f)
cos x — cos3x
Vš
(g) cosacos2x = cos4jccos5a"
(h) sin3jc = 2 siru
(i) tgx — sin.v +cosjc = 1
(j) 2sin2A" + cosjc = 2sin2A" cos.t + 1
(k) J^+^L = 1 l-VŠtgx
3
(1) sin2 2x + sin2 4x = -
2
(m) |sinjc| = sinjc + 2
(n) |tgx + cotgx|
Vš
K= U {ln + kn, 2kx, \n + 2kn}
keZ
K= U {f + ^f}
keZ
K= U {f + kn}
keZ,
K= u {^f}
keZ
K= U {km, %+kn, \n + kn]
keZ
K= U {2k%, f + k7t}
kel
K= U {f +k%,2kn}
kel
K= U {{^r + ta}
* = u {f+*f,t+*!,f+*f}
£e2
a:= u {|tt+2^}
AreZ
^= U {ff +     %n + kn, žn + kn]
kel