5.4
Page Section 1
Example 5.4.23 Solve cos 2x + cos x = 0 algebraically for exact solutions in the interval [0, 2). cos 2x + cos x = cos2 x - sin2 x + cos x = cos2 x - (1 - cos2 x) + cos x = 2 cos2 x + cos x - 1 = 0 Let y = cos x. Then cos 2x + cos x = 2 cos2 x + cos x - 1 = 0 = 2y 2 + y - 1 = 0 -b b2 - 4ac y = 2a -1 1 + 8 = 4 -1 3 = 4 -4 2 or = 4 4 1 = or - 1 2 So we must solve y = cos x = 1/2 and y = cos x = -1. The equation cos x = -1 has a solution of in the interval [0, 2). The equation cos x = adj/hyp = 1/2 corresponds to one of our special triangles:
hyp=2 /3 adj=1
opp= 3
So the solution is /3. There is also a solution at 2 - /3 = 5/3 in the interval [0, 2) (the solution in Quadrant IV). The solutions to cos 2x + cos x = 0 in the interval [0, 2) are 5 , , . 3 3
Section 5.4
Page 2
Example 5.4.24 Solve cos 2x + sin x = 0 algebraically for exact solutions the in interval [0, 2). cos 2x + sin x = cos2 x - sin2 x + sin x = 1 - sin2 x - sin2 x + sin x = -2 sin2 x + sin x + 1 = 0 Let y = sin x. Then cos 2x + sin x = -2 sin2 x + sin x + 1 = 0 = -2y 2 + y + 1 = 0 -b b2 - 4ac y = 2a -1 1 + 8 -1 3 = = -4 -4 2 -4 = or -4 -4 1 = - or 1 2 So we must solve y = sin x = -1/2 and y = sin x = 1. The equation sin x = 1 has a solution of /2 in the interval [0, 2). The equation sin x = opp/hyp = 1/2 corresponds to one of our special triangles:
hyp=2 /6 adj= 3 opp=1
So the solution ...