(tan x+1)(cos x+1)=0: solve on the interval [0,2π]

[ravenschyld]

(tan x+1)(cos x+1)=0: solve on the interval [0,2π]

I am not sure how to do this. I know the answer is 3π/4, π, and 7π/4, but I am at a complete loss. Thanks for your help.

 

[pka]

I am not sure how to do this. I know the answer is 3π/4, π, and 7π/4, but I am at a complete loss.

WHY is that?
Can you solve \(\displaystyle \tan(x)=-1~?\)

Can you solve \(\displaystyle \cos(x)=-1~?\)

If not, why are you asked to?

 

[ravenschyld]

WHY is that?

Truth be told, I am having a very hard time grasping the trig equations.

tan x = -1
x=arctan (-1)
x=-π/4 + πk; 3π/4 + πk

cos x = -1
x=arccos (-1)
x=π + 2πk

Is this correct?

 

[pka]

Truth be told, I am having a very hard time grasping the trig equations.
tan x = -1
x=arctan (-1)
x=-π/4 + πk; 3π/4 + πk

cos x = -1
x=arccos (-1)
x=π + 2πk

Actually \(\displaystyle \arctan~\&~\arccos\) really are not needed.
You should have memorized all values for \(\displaystyle 0,\dfrac{\pi}{6},\dfrac{\pi}{4},\dfrac{\pi}{3}, \dfrac{\pi}{2}\) for \(\displaystyle \sin,~cos,~\&~\tan\)

Do it and your life will be so much easier.