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Solve sin2x + 4sin x + 3 = 0

Mathematics
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do u know how to solve this ?
sin(2x) is more interesting :)
Is it sin^2x + 4sinx + 3 = 0 ?

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Other answers:

|dw:1345778720492:dw| i dont know why latex is not working..
@Lulu212 Is it \(\sin 2x \ or \ \sin^2 x\)?
sin 2x
\[sin2x + 4sinx + 3\]
\[sin2x = 2sinxcosx\] \[ 2sinxcosx + 4sinx + 3 = 0 \]
have you tried that yet?
hm...hm.... u need...to find the General soln of this....!
\[ \sin(2x) + 4\sin (x) + 3 = 0\] \[ 2\sin(x)\cos(x) + 4\sin (x) + 3 = 0\] \[2\sin (x)\left(\cos(x)+2\right)+3=0\] \[2\sin (x)\left(\cos(x)+2\right)=-3\] \[\cos(x)+2=\frac{-3}2\csc(x)\] i dont know
I don't understand why @UnkleRhaukus can't finish it
Can u Finish @Hero i am also stuck..there
hmm http://www.wolframalpha.com/input/?i=sin2x+%2B+4sin+x+%2B+3+%3D+0 i dont think its possible..
i think..tthere is mistake in question..))))
@Lulu212 can u check ur question...))
I think it should be \(sin^2x + 4sinx +3 =0\)
then \(u=sinx\) more reasonable
then it is a quad eq
we can find sinx
I'm inclined to agree with @Mimi_x3
this question can be solved ............why u guys are worrying????????
it cant be solved just endless ways of writing the equation
what are we solving for? LOL. that equation needs to be written precisely in a way that we all can agree to what it is.
\[\sin^2x+4\sin x+3=0\] \[u^2-4u+3=0\] \[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\] \[u=2\pm\frac{\sqrt{4}}{2}\]\[u=2\pm1\]\[u=1,3\] \[(1)^2+4(1)+3=0,\qquad\qquad 3^2-4(3)+3=0\]\[8=0,\qquad\qquad\qquad 3^2-4(3)+3=0\]\[\text{false}\qquad\qquad\qquad 9-12+3=0\]\[\qquad\qquad\qquad\qquad0=0\]\[\qquad\qquad\qquad\text{true}\]
the equation toolbar in this thing sucks hairy ______. how do you write it so neatly like that?
so \(u=\sin(x)=3\) \[x=\arcsin (3)\] ... this question is very confusing \[\text{\[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\]}\]
oh damn my solution is way off \[\sin^2(\arcsin 3) + 4\sin (\arcsin 3) + 3 = 24≠0\]
oh \(u^2+4u+3=0\) NOT \(u^2-4u+3=0\)
but this wasnt the problem that was stated lol
\[u_{1,2}=-2\pm1=-3,-1\]
\[\sin x=u=-3,-1\] \[x_{1,2}=\arcsin (-1),\quad \arcsin(-3)\] \[x_1=-\frac \pi2\qquad, x_2=\text{some transcendental nonsense} \] hence\[x=-\frac {\pi}2\] CHECKING \[\sin^2x + 4\sin x + 3 = 0\] \[\sin^2\left(-\frac\pi 2\right) + 4\sin \left(-\frac\pi 2\right) + 3 = 0\]\[1+4(-1)+3=0\]\[0=0\]true hence \[\large{x=-\frac {\pi}2}\]is the solution to \[\sin^2x + 4\sin x + 3 = 0\]
ta-da
People who post questions need to learn how to use the equation editors to avoid such confusion.
couldn't agree more @Hero
i think someone needs to write a program that can interpet questions and convert them to \(\LaTeX\) automagically
@lulu212 i am disappoint at this equation
I imagine that something like that would have to be interactive. It would have to be some kind of AI that would automatically ask the user something like: Did you mean \(\sin(2x)\) or \(\sin^2(x)\)?
beyond OS' capacity :P we can ask Wolfram though. That dude is probably the smartest programmer I've seen alive
That's because he was reincarnated Einstein
hahaha... I think he was already born when Einstein was still alive
Well, Einstein's lost twin
this question has like 40 responses all because of a typing error
We're still assuming that it is a typing error
it is
There's no such thing as typing errors. Only human errors.
the person who asked for help did not even say anything after the question was posted (TYPOGRAPHICAL ERROR)
can't we all be einsteins lost twin?
I know that I can't... I have a roach's brain compare to his
same difference hero lol
twins dont necessarily have the same charistics
oh that's right... I'd be the opposite of his intellect then
* characteristic
In one of Einstein's lost journals, he wrote that his brain was removed as a kid. The person handling his brain accidentally dropped in brain in a toilet, but quietly removed it without telling anyone. Once the brain was placed back into Einstein's head, he felt different, like as if he had more 'powers' and 'ability'. We all know what that led to.
oh my.
where's David Hume?

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