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Solve tan 2x - cot x = 0

Solve tan 2x - cot x = 0 for the interval [0, 2pi)

Comments

Write tan2x as sin2x/cos2x and cots as cosx/Sinx.
Problem becomes:
2sinxcosx/(cos^2 x - sin^2 x) - cosx/sinx = 0. 
Get a common denominator for this difference,
combine like terms and set numerator to 0. 
This yields cosx(3sin^2 x - cos^2 x) = 0.  Set each factor to 0 and you get pi/2, 3pi/2, pi/6 and 7pi/6 as your answers. 

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Kevin C. | Successful Math Tutor -- Recently retired high school math teacherSuccessful Math Tutor -- Recently retire...
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Use the following identities:  cot x = 1/tanx. and tan 2x = (2tanx)/(1-tan2x)
 
The equation then becomes (2tanx)/(1-tan2x) - 1/(tanx) = 0
 
Add 1/(tanx) to each side:  (2tanx)/(1-tan2x) = 1/(tanx).
 
Multiply each side by tanx*(1-tan2x):  2tan2x = 1-tan2x.
 
Combine terms:  3tan2x = 1,  divide by 3:  tan2x = 3,  so tanx = √3 and tanx = -√3.
 
so x = π/3 in each of the 4 quadrants.

Comments

AAARG!  I don't believe I made that mistake.
 
tanx = 1/√3 and x = -1/√3  so x = π/6 in each of the 4 quadrants.  Sorry.
Wrong answer. Tan^2(x) should equal 1/3 not 3!!
As I said in my correction above.
This solution misses two additional solutions: π/2 and 3π/2.  Why did that happen?