Question

How do you solve `cosx tanx - 2 cos^2x = -1` on the interval `[0,2pi]`?

Answer 1

Solve: cos x.tan x - 2cos^2 x = - 1

Explanation:

`cos x(sin x /cos x) = 2cos^2 x - 1`
sin x = cos 2x (Condition cos x not zero, x not `(pi)/2` and `(3pi)/2`)
`sin x = 1 - 2sin^2 x`. Call sin x = t. We get quadratic equation:

`2t^2 + t - 1 = 0`
(a - b + c) = 0 --> 2 real roots: t = -1 and t = -c/a = 1/2

a. t = sin x = - 1 --> `x = (3pi)/2` (Rejected)

b. `t = sin x = 1/2` --> `x = (pi)/6`and `x = (5pi)/6`
Check.
`x = (pi)/6` --> `cos x = sqrt3/2` -->`tan x = sqrt3/3` -->
`cos^2 x = 3/4`

`(sqrt3/2)((sqrt3)/3) - 2(3/4) = 3/6 - 3/2 = -6/6 = - 1. OK`