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rsnoobBest ResponseYou've already chosen the best response.0
For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x  sin^2 x = cos^2 x  (1cos^2) = 2cos^2 x  1. So we have (2cos^2 x  1)(2cos x + 1) = 0. So 2cos^2 x 1 = 0 or x = 0 and 2pi. For 2sec^2 x + tan^2 x  3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have 2(tan^2 x + 1) + tan^2 x  3 = 0 or 2tan^2 x + tan^2 x  1 = 0 or 3 tan^2 x = 1. So x = pi/2, pi/2 + pi = 3pi/2.
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
hi the first thing I tried to solve this problem was to factor it so I have (cos x+1)(cos x+1) but now I dont know what to next
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Unclear how to proceed. What is the first term, \(\cos(2x)\) or \(\cos^{2}(x)\)?
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
cos \[\cos^{2}x\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Perfect. The we have only to solve  by factoring. \(\cos^{2}x + 2\cos x + 1 = 0\) \((\cos(x) + 1)^{2} = 0\) Now what?
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
That is where I got stuck I am not sure what to do after factoring
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
What's the point of factoring? Why do we do that at all? If a*b = 0, what do we know about a or b? To contract, if a*b = 4, what do we NOT know about either a or b?
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
what each of the equal?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Kind of. a*b = 0 tells us that either a = 0 or b = 0. What can we tell about a and b when it isn't zero? a*b = 4, for example? Is one of them zero? Aboslutely not, but that's about all we can tell. Is a = 4? Maybe. Is b = 12? Maybe. Not a clue. Only "=0" is particularly helpful. So, what should we do with that factored expression?
 one year ago

chaguanasBest ResponseYou've already chosen the best response.0
tkhunny your explanations are more complex than the problem itself :)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
That's why we invent notation, so that problems can be simpler than explanations. If we have this: Factor1 * Factor2 = 0 Then we must have Either this Factor1 = 0 or this Factor2 = 0 How can we apply this to your factored equation?
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
cos x+1(0)=0 I am not sure
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
\((\cos(x)+1)^{2} = 0\) We have two factors, but hey are exactly the same. This limits the possible results. \(\cos(x) + 1 = 0\) \(\cos(x) = 1\) \(x = \dfrac{3\pi}{2} + 2k\pi\) where \(k\in \mathbb{Z}\) Do you know what all that means?
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
I do... but if cos(x)=1 wouldn't [x=\pi\]
 one year ago

marasofia1616Best ResponseYou've already chosen the best response.1
anyway I got it ... thanks for the help!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Awesome. How did I do that?! Right you are. \(x = \pi + 2k\pi\)
 one year ago
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