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