SOLUTION: The solution(s) of the equation 2cos^2 x tan x=tan x, over the interval [0,2π) is (are)? Can you please explain to me how you got the answer? (a) x=0 (b) &#119909

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The solution(s) of the equation 2cos^2 x tan x=tan x, over the interval [0,2π) is (are)? Can you please explain to me how you got the answer? (a) x=0 (b) &#119909      Log On


   

Question 761442: The solution(s) of the equation
2cos^2 x tan x=tan x, over the interval [0,2π)
is (are)?
Can you please explain to me how you got the answer?
(a) x=0
(b) 𝑥=0,𝜋
(c) 𝑥=0,𝜋4,3𝜋4,𝜋,5𝜋4
(d) 𝑥=0,𝜋4,3𝜋4,𝜋,5𝜋4,7𝜋4
(e) None of the above.
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
2cos^2xtanx = tanx
2cos^2xtanx - tanx = 0
tanx(2cos^2x - 1) = 0
tanx = 0
x = 0, pi, 2pi.
2cos^x - 1 = 0
2cos^2x = 1
cos^2x = 1/2
cosx = 1/sqrt2
x = pi/4, 7pi/4
[0,pi/4,pi,7pi/4 and 2pi]
Hope this helps.
:-)