Trig help...again :)
\[2\sin \theta \cos \theta + \sqrt{3} \cos \theta = 0\]
make some attempts
..I just need help getting started....I don't like direct answers @hwyl, so I'm not asking for one....
like what kind of help to get started? you said "again," so it means you are not doing this for the first time. What part of the problem you don't understand?
I meant I'm not doing trig for the first time....this is the first time that I am doing this...
Not all of my trig is like this....not much is actually.
first, clarify if the second term of the problem is \(\sqrt{3}~cos \theta \) or \(\sqrt{3cos\theta} \)
It's what I put...\[\sqrt {3}~ cos \theta\]
I want to make sure because people make mistakes putting it in latex form.
:)
is there anything that is factorable?
What do you mean? O.o
can you factor the left-hand-side (LHS)?
\(ab + bc \rightarrow b(a+c) \)
in my own opinion, factoring simplifies and does many wonders.
I'm not quite sure how to factor this....
well look at the example I showed you look for something common in the first and second term. from what I can see, \(cos ~\theta \)
So \[cos (2 sin + \sqrt3)?\]
perfect! now complete the form \(ab + bc = 0 \rightarrow b(a+c) = 0 \rightarrow b =0, ~ a+c = 0 \) what we are basically looking for are the values when the curves or lines passes through the x-axis, which we call the ZEROES.
so find the values where cos is 0?
and where 2 sin + sqrt 3 equal 0?
if so, then 2 of them are pi/2 and 3pi/2. How do I find the other ones? I'm not quite sure how to do it with the 2....
you're on the right track... you can do on your own
Wait! 4pi/3?
and 5pi/3?
easy peachy, isn't it?
yes! thank you!
you did well and I think you're just going to need a few algebra techniques to get started.
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