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beknazar23
 one year ago
Best ResponseYou've already chosen the best response.0sinx(sinx+2cosx) = 0 What can you do now? think.

beknazar23
 one year ago
Best ResponseYou've already chosen the best response.0if a*b = 0, Either a = 0 or b = 0 Here, a = sinx and b = sinx + 2cosx

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0i don't think this is that easy

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0that 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 \[\sin(x)=2\cos(x)\]

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0that is not to say you cannot do it, but it is going to take some work

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0the first thing you have to do to solve \[\sin(x)+2\cos(x)=0\] is to write it as a single function of sine

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0in general you can write \[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\]

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0in this case you get \[\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)\] where \(\tan(\theta)=2\)

beknazar23
 one year ago
Best ResponseYou've already chosen the best response.0tanx = 2 x = atan(2) + n*(pi)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0therefore \(x+\theta=n\pi\) gives \(\theta=n\pi\tan^{1}(2)\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0you can check that it is right by looking here http://www.wolframalpha.com/input/?i=sin%28x%29%2B2cos%28x%29

satellite73
 one year ago
Best 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

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

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