Question

Find the solution {in the interval of [0,2pi)} of:

tan2x - 2cosx = 0

( I know the answer is pi/6, but I guessed and checked. Please show the steps I need to go through in order to get the answer )

Answer 1

tan 2x = sin 2x / cos 2x
= 2 sin x cos x /(cos^2 x - sin^2 x)
= 2 cos x [ sin x / (cos^2 x-sin^2 x) ]
now if we can find where
[ sin x / (cos^2x-sin^2x) ]
is 1
then we have it
well cos^2 x = 1 - sin^2 x
so we have
sin x /[ 1 - sin^2 x - sin^2 x] = 1
or
sin x = 1 - 2 sin^2 x
2 sin^2 x + sin x -1 = 0
(2 sin x -1) (sin x + 1) = 0
sin x = -1
or
sin x = 1/2
if x = 30 degrees (pi/6 radians) that is a solution
so is x = 180 - 30 = 150 degrees = 5pi/6
also 3 pi/4 gives x = -1

Answer 2

tan 2x = 2 tan x/[1 - tan^2x] = 2 cos x
tan x = cos x - cos x tan^2 x
= cos x - sin^2 x/cos x
Multiply both sides by cos x.
sin x = cos^2 x - sin^2 x = cos 2x
= 1 - 2 sin^2 x
2 sin^2 x + sin x -1 = 0
(2 sin x -1)(sin x + 1) = 0
sin x = 1/2 or -1
x = pi/6 or 3 pi/2

There are two answers

Answer 3

also 3 pi/2 (270 degrees) gives x = -1

Answer 4

Thank you!