SOLUTION: Hi I have tried this problem a few times. I was able to get half of the problem somehow, but I can't get the last 2 answers.
4tan(theta)^2cos(theta)-2tan(theta)=0
I found hal
Algebra ->
Trigonometry-basics
-> SOLUTION: Hi I have tried this problem a few times. I was able to get half of the problem somehow, but I can't get the last 2 answers.
4tan(theta)^2cos(theta)-2tan(theta)=0
I found hal
Log On
Question 1007689: Hi I have tried this problem a few times. I was able to get half of the problem somehow, but I can't get the last 2 answers.
4tan(theta)^2cos(theta)-2tan(theta)=0
I found half the answer by dividing it all by 2tan(theta)^2 which turns into
2cos(theta)=1 and then
cos = 1/2. I am still missing another part.
Thanks Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 4tan(theta)^2cos(theta)-2tan(theta)=0
I found half the answer by dividing it all by 2tan(theta)^2 which turns into
2cos(theta)=1 and then
cos = 1/2. I am still missing another part.
---------------------------------------
4tan(theta)^2cos(theta)-2tan(theta)=0
Factor::
2tan(t)[2tan(t)cos(t) - 1) = 0
-----
2tan(t)[2sin(t)-1] = 0
-----
tan(t) = 0 or sin(t) = 1/2
-------
t = 0 or pi OR t = pi/6 or (5/6)pi
-------------
Cheers,
Stan H.
--------------
(