6 + 2 sen x = 1 2 cos2x – cos x = 0 2 cos2x + 5 sen x = –1

Presentación del tema: "6 + 2 sen x = 1 2 cos2x – cos x = 0 2 cos2x + 5 sen x = –1"— Transcripción de la presentación:

1 6 + 2 sen x = 1 2 cos2x – cos x = 0 2 cos2x + 5 sen x = –1
Ejercicios sobre Ecuaciones trigonométricas 2 cos2x – cos x = 0 2 cos2x + 5 sen x = –1

2 sen 2x = tan x Revisión del estudio individual C) sen x cosx
2 senxcos2x = senx 2 senx (1 – sen2x) – sen x = 0 2 senx – 2 sen3x – sen x = 0 2 sen3x – sen x = 0 senx(2sen2x – 1) = 0

3 senx(2sen2x – 1) = 0 1 2 sen2x = sen x = 0 ó senx =  2 2 4  x = k ó x = + k 2 kZ

4 Ejercicio Resuelve las siguientes ecuaciones:
a) 2 sen 2x cot x – cos 2x = 3 cos x sen 2x cos x b) = 2 sen2x sen x c) sen 2x = 2 – 2 cos 2x

5  x1= x3= 2k  x2= 2 – = a) 2 sen 2x cot x – cos 2x = 3 cos x cosx
4senx cosx –(2cos2x – 1) = 3 cos x 4 cos2x – 2 cos2x + 1 = 3 cos x 2 cos2x – 3 cos x + 1 = 0 (2 cos x – 1)(cos x – 1) = 0 1 2 cos x = ó cos x = 1  3 x1= + 2k ; k x3= 2k  3 5 3 = x2= 2 – + 2k ; k

6 k   sen 2x cos x b) = 2 sen2x sen x 2 senx cosx cosx = 2 sen2x sen x
: 2 cos2x – sen2x = 0 cos 2x = 0 k    2 3 2 2x1 = + 2k 2x2 = + 2k x1 =  4 + k 3 4 x2 = + k

7 c) sen 2x = 2 – 2 cos 2x : 2 2 senx cosx = 2 (1 – cos2x) senx cosx = [1 – (1 – 2sen2x)] senx cosx = [1 – 1 + 2sen2x)] senx cosx = 2 sen2x 2 sen2x – senx cosx = 0 senx (2 senx – cosx) = 0

8     5 senx (2 senx – cosx)= 0 sen x = 0 ó 2 senx – cosx = 0
x = 1800k 2 senx = cosx kZ 4sen2x = cos2x 4sen2x = 1 – sen2x 5sen2x = 1 sen x =  1 5  sen x = 5  5 

9 k  Z   5 TABLA sen x = 5 sen x =  0,448 x1 = 26,60 + 3600k

10 = a) sen2x + cosx = 0 b) cos2x + cos2x = 5sen2x
Para el estudio individual Resuelve las siguientes ecuaciones con la condición 0o  x  360o . a) sen2x + cosx = 0 x = 90o; 270o; 210o; 330o b) cos2x + cos2x = 5sen2x x = 30o; 150o ;210o ;330o c) cos2x + cosx = 0 x=60o; 300o; 180o cos2x + sen2x 1 – senx = 5 3 d) x = 90o; 41,8o; 138,2o