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beknazar23Best ResponseYou've already chosen the best response.0
sinx(sinx+2cosx) = 0 What can you do now? think.
 one year ago

beknazar23Best ResponseYou've already chosen the best response.0
if a*b = 0, Either a = 0 or b = 0 Here, a = sinx and b = sinx + 2cosx
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i don't think this is that easy
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
that is \(\sin(x)=0\) is easy enough to solve, you get \(x=n\pi, n\in \mathbb{Z}\) but i can't see a good way to solve \[\sin(x)=2\cos(x)\]
 one year ago

TheViperBest ResponseYou've already chosen the best response.0
\[sin^2x+2\ sin\ x\ cos\ x=0\] \[sin\ x(sin\ x+2\ cos\ x)=0\] \[(sin\ x+2\ cos\ x)=0\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
that is not to say you cannot do it, but it is going to take some work
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
the first thing you have to do to solve \[\sin(x)+2\cos(x)=0\] is to write it as a single function of sine
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
in general you can write \[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\]
 one year ago

tayluvsyaBest ResponseYou've already chosen the best response.0
:'( i need help on mine
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I kinda wonder if you could do \(\sin^2(x) + \sin(2x)=0\).
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
in this case you get \[\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)\] where \(\tan(\theta)=2\)
 one year ago

beknazar23Best ResponseYou've already chosen the best response.0
tanx = 2 x = atan(2) + n*(pi)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
therefore \(x+\theta=n\pi\) gives \(\theta=n\pi\tan^{1}(2)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you can check that it is right by looking here http://www.wolframalpha.com/input/?i=sin%28x%29%2B2cos%28x%29
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
@ParthKohli i don't think that identity is going to help in this case you will have two arguments, one of \(x\) and the other of \(2x\) if you have a solution from that approach i would love to see it
 one year ago

TheViperBest ResponseYou've already chosen the best response.0
oh I had forgotten this site Thanx @satellite73 for remembering :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Gaahh @satellite73 you were right.
 one year ago
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