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Hmm...
\[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]

after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]

2cosθsinθ = sin2θ
Divide both sides by 2
cosθsinθ = ...?

\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]

Multiply both sides by sin2θ, what do you get?

wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]

The first one.

wait, haha I'm sorry!...

its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]

Next:
Use
cos2x = 1-2(sinx)^2

so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]

riiiight! I see it.

\[cos2x = 1-2sin^2x\]\[2sin^2x = 1-cos2x\]Replace x by 2θ. Life isn't difficult then.

It should be\[\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta\]

i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]

Are you sure...???

\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

2theta =pi/4, 3pi/4

so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??

yupp

now, how do I find the the equation in the interval [0, 2π)??

:)

wow! at the same time

ohhhh ok

But it's up to you :)

From zero to pi, yes... keep going to 2pi... 4 more solutions.

What have you put?

pi/8, 3pi/8 my first try
and pi/8, 3pi/8, 5pi/8, 7pi/8 the second try. I have one more try :S

I don't like to just give answers, but you are working very hard at this. ;)

its [0, 2pi)

ok. hahahah thank you so sooooo much! and I'm sorry for the trouble

ok. thanks!

Agree with @EulerGroupie 's saying :)