Goniometrické rovnice
(Seminář z matematiky i - m1130/02 2015)
(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*- ^ = 0
(e) 3 sin* = 2cos2*
(f) sin* + cos2* = -
4
(g) 3cos2* — 4cos* — sin2* — 2 = 0
(h) cos2* + sin* = 0
(i) sin* = cos*
(j) sin*+ \/3 cos* = 0 /i x sin-*
(k) —=■ + cos* = 1 V3
(1) y/3sin* + cos* = v7^
K= U {ta, f + 2ta, |^: + 2^}
* = U {f+2Jbr}
fceZ
* = U {f+ta, jn + kn}
fceZ
a:= u {f+ 2ta, |^+2^}
fceZ
kel
K= U {iTZ + lkTZ^TZ + lkTt}
kel
kel
K= U {f+ ^}
* = U {j^ + ta}
a:= u {2ta,f+2ta}
fceZ
a:= U {f+ 2ta,f+2Jbr}
feeZ
(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 {Jbr,f+ *f}
feeZ
a:= u {§+Jbr, § + 2ta, §rc+2Jbr}
* = U {f +ta, §rc + ta, f + ta}
feeZ
(d) cosA + sinx
cos 2x
1 — sinŽA" (e) cos4jc — sin4jc = sin2jc sku + sin3jc
(f)
cosx — cos 3.x:
V3
(g) cosacos2x = cos4jccos5a"
(h) sin3jc = 2sinjc
(i) tgx — sinx + cosjc = 1
(j) 2 sin2.* + cos a = 2sin2a" cos.t + 1 l-VŠtgjc
(1) sin2 2x + sin2 4x = -
2
(m) |sinjc| = shut+ 2
(n) |tgjc + cotgjc|
V3
K= U {\n + kK, 2kn, \n + 2kn}
K= U {f + £f}
K= U {% + kn}
k<EZ
K= u {£f}
fee
^= U {f +k%,2kn}
K= U +
* = U {f +*f,§ + *#,f + £f}
£■ = U {\% + Zk%}
keZ