Solve the equation 2 cos x - sec x = 1 on the interval [0, 2π). how would i do this

Guest Mar 25, 2015

#1**+5 **

2 cos x - sec x = 1 .....note sec x = 1/cosx so we have

2cosx - 1/cosx = 1 multiply through by cos x

2cos^2 x - 1 = cos x rearrange

2cos^2 x - cos x - 1 = 0 factor

(2 cos x + 1)(cos x - 1) = 0 so either

2 cos x + 1 = 0

2 cos x = -1

cos x = -1//2 and this occurs at 2pi/3 and 4pi/3

OR

cosx - 1 = 0

cos x = 1 and this occurs at 0

So the solutions on the given interval are 0, 2pi/3 and 4pi/3

CPhill
Mar 25, 2015

#1**+5 **

Best Answer

2 cos x - sec x = 1 .....note sec x = 1/cosx so we have

2cosx - 1/cosx = 1 multiply through by cos x

2cos^2 x - 1 = cos x rearrange

2cos^2 x - cos x - 1 = 0 factor

(2 cos x + 1)(cos x - 1) = 0 so either

2 cos x + 1 = 0

2 cos x = -1

cos x = -1//2 and this occurs at 2pi/3 and 4pi/3

OR

cosx - 1 = 0

cos x = 1 and this occurs at 0

So the solutions on the given interval are 0, 2pi/3 and 4pi/3

CPhill
Mar 25, 2015