How do you find the exact solutions of the equation #tan2x-2cosx=0# in the interval #[0,2pi)#?

1 Answer
Jan 5, 2017

#x = pi/6, (5pi)/6 (3pi)/2#

Explanation:

Recall that #tanx# can be written as #sinx/cosx#:

#(sin2x)/(cos2x) = 2cosx#

#sin2x = 2cosx(cos2x)#

#(sin2x)/(2cosx) = cos2x#

#sin2x# can be written as #2sinxcosx#

#(2sinxcosx)/(2cosx) = cos2x#

#sinx= cos2x#

#cos2x# can be expanded as #cos^2x - sin^2x#.

#sinx = cos^2x- sin^2x#

#sinx = 1 - sin^2x - sin^2x#

#sinx = 1 - 2sin^2x#

#2sin^2x + sinx - 1 = 0#

Let #t = sinx#:

#2t^2 + t - 1 = 0#

#2t^2 + 2t - t - 1 = 0#

#2t(t + 1) - (t + 1) =0#

#(2t - 1)(t + 1) = 0#

#t = 1/2 and -1#

#sinx = 1/2 and sinx= -1#

#x = pi/6, (5pi)/6 (3pi)/2#

Hopefully this helps!