Solve cos2t+cost=0 for 0

Question

Solve cos2t+cost=0 for 0<t<360 degrees
please explain how to do it too, thanks!
i got 90 degrees idk what to do
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anonymous · Answer

$$cos(2t) = cos^2t-sin^2t = 1- 2sin^2t$$
$$cos(t)+cos^2t-sin^2t = 0$$
$$cos(t)+1- 2sin^2t = 0$$
$$cos(t) = 2sin^2t - 1$$
$$cos(t) = -1(1-2sin^2t)$$

I'm just playing around with the double-angle formula hoping to see something.  I don't see it, but maybe you will...  Sorry, I can't help with this much more...

anonymous · Answer

Here is the answer though.  I am not sure how they arrived at it.

anonymous · Answer

Is that cos(2t) or $$cos^2t$$
?  That will make a big difference; just checking...

anonymous · Answer

its is cos 2t...not cos^2 t

anonymous · Answer

it*

anonymous · Answer

Okay.  Well, since I'm not sure exactly how to solve this, if I were presented this on an exam I would draw my unit circle and plug in values for t that would make cos(2t) + cos(t) = 0.  Just a fallback plan if you don't figure this out (although I don't advocate relying on this long-term).

anonymous · Answer

is sint+cost=1 a legitimate pythagorean/trig identity?

anonymous · Answer

perhaps i could use that?

anonymous · Answer

sin^2(x) + cos^2(x) = 1

anonymous · Answer

but if you square root it, could it work?

anonymous · Answer

You'd be taking the square-root of the whole left side which is different than sinx + cosx

anonymous · Answer

This might help: It used it religiously back in trig and cal2. http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

anonymous · Answer

(although I never really spent enough time memorizing the trig identities..)

anonymous · Answer

alright, going to bed.  good luck.  try tagging some moderators or level-99 users with @username, like @mathbrz

anonymous · Answer

you can find them listed on the main page if they're online

anonymous · Answer

okay thanks

aravindg · Answer

Looks like your qn is already answered?

anonymous · Answer

no i dont have an answer yet.

anonymous · Answer

we were working towrds an answer, trying to explore our options

anonymous · Answer

i used the pythagorean identity to get 90 degrees and 270 degrees but turns out that i used it incorrectly.

anonymous · Answer

alright well im going to bed. it is late and i havent done my homework and tomorrow is going to be hard. if anyone figures out how to do this please post answers and explanations or hints on how to solve. thank you.