Which value of x in the interval -π/2 < x < π/2 satisfies the equation cos(x) = sec(x)?

cos(x) = sec(x)
cos(x) = 1/cos(x)
cos²(x) - 1 = 0
[cos(x) - 1] [cos(x) + 1] = 0
cos(x) = 1  or  cos(x) = -1
x = 0          or  x has no solution in -π/2 < x < π/2


The answer: C) 0

secx = 1/cosx

so, cosx = 1/cosx

=> cos²x = 1

i.e. cosx = ±1

so, x =  ±2nπ or ±π + 2nπ 

With n = 0 we get x = 0 and ±π

Hence, x = 0 is the only solution for -π/2 < x < π/2

:)>

Since sec = 1/cos, equation cos = sec is equivalent to equation cos = 1/cos, ie.,
cos^2 = 1. In (-pi/2,pi/2) this occurs only when x = 0. C) gives correct answer.