A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Solve sin2x + 4sin x + 3 = 0
anonymous
 3 years ago
Solve sin2x + 4sin x + 3 = 0

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do u know how to solve this ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin(2x) is more interesting :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it sin^2x + 4sinx + 3 = 0 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345778720492:dw i dont know why latex is not working..

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0@Lulu212 Is it \(\sin 2x \ or \ \sin^2 x\)?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1\[sin2x = 2sinxcosx\] \[ 2sinxcosx + 4sinx + 3 = 0 \]

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1have you tried that yet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hm...hm.... u need...to find the General soln of this....!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @Hero @jim_thompson5910 @myininaya

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \sin(2x) + 4\sin (x) + 3 = 0\] \[ 2\sin(x)\cos(x) + 4\sin (x) + 3 = 0\] \[2\sin (x)\left(\cos(x)+2\right)+3=0\] \[2\sin (x)\left(\cos(x)+2\right)=3\] \[\cos(x)+2=\frac{3}2\csc(x)\] i dont know

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1I don't understand why @UnkleRhaukus can't finish it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can u Finish @Hero i am also stuck..there

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1hmm http://www.wolframalpha.com/input/?i=sin2x+%2B+4sin+x+%2B+3+%3D+0 i dont think its possible..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think..tthere is mistake in question..))))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Lulu212 can u check ur question...))

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1I think it should be \(sin^2x + 4sinx +3 =0\)

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1then \(u=sinx\) more reasonable

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1I'm inclined to agree with @Mimi_x3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this question can be solved ............why u guys are worrying????????

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it cant be solved just endless ways of writing the equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what are we solving for? LOL. that equation needs to be written precisely in a way that we all can agree to what it is.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sin^2x+4\sin x+3=0\] \[u^24u+3=0\] \[u=\frac{(4)\pm\sqrt{(4)^24\times(1)\times(3)}}{2(1)}\] \[u=2\pm\frac{\sqrt{4}}{2}\]\[u=2\pm1\]\[u=1,3\] \[(1)^2+4(1)+3=0,\qquad\qquad 3^24(3)+3=0\]\[8=0,\qquad\qquad\qquad 3^24(3)+3=0\]\[\text{false}\qquad\qquad\qquad 912+3=0\]\[\qquad\qquad\qquad\qquad0=0\]\[\qquad\qquad\qquad\text{true}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the equation toolbar in this thing sucks hairy ______. how do you write it so neatly like that?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1so \(u=\sin(x)=3\) \[x=\arcsin (3)\] ... this question is very confusing \[\text{\[u=\frac{(4)\pm\sqrt{(4)^24\times(1)\times(3)}}{2(1)}\]}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1oh damn my solution is way off \[\sin^2(\arcsin 3) + 4\sin (\arcsin 3) + 3 = 24≠0\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1oh \(u^2+4u+3=0\) NOT \(u^24u+3=0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but this wasnt the problem that was stated lol

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[u_{1,2}=2\pm1=3,1\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.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 \[\sin^2x + 4\sin x + 3 = 0\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1People who post questions need to learn how to use the equation editors to avoid such confusion.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0couldn't agree more @Hero

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i think someone needs to write a program that can interpet questions and convert them to \(\LaTeX\) automagically

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@lulu212 i am disappoint at this equation

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1I 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 \(\sin^2(x)\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0beyond OS' capacity :P we can ask Wolfram though. That dude is probably the smartest programmer I've seen alive

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1That's because he was reincarnated Einstein

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hahaha... I think he was already born when Einstein was still alive

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1Well, Einstein's lost twin

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this question has like 40 responses all because of a typing error

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1We're still assuming that it is a typing error

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1There's no such thing as typing errors. Only human errors.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the person who asked for help did not even say anything after the question was posted (TYPOGRAPHICAL ERROR)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1can't we all be einsteins lost twin?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know that I can't... I have a roach's brain compare to his

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0same difference hero lol

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1twins dont necessarily have the same charistics

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh that's right... I'd be the opposite of his intellect then

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1In 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.