Solve sin2x + 4sin x + 3 = 0

- anonymous

Solve sin2x + 4sin x + 3 = 0

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

do u know how to solve this ?

- anonymous

sin(2x) is more interesting :)

- anonymous

Is it sin^2x + 4sinx + 3 = 0 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

|dw:1345778720492:dw|
i dont know why latex is not working..

- ash2326

@Lulu212 Is it
\(\sin 2x \ or \ \sin^2 x\)?

- anonymous

sin 2x

- Mimi_x3

\[sin2x + 4sinx + 3\]

- Mimi_x3

\[sin2x = 2sinxcosx\]
\[ 2sinxcosx + 4sinx + 3 = 0 \]

- Mimi_x3

have you tried that yet?

- anonymous

hm...hm.... u need...to find the General soln of this....!

- anonymous

@UnkleRhaukus @Hero @jim_thompson5910 @myininaya

- anonymous

@Hero @hartnn

- UnkleRhaukus

\[ \sin(2x) + 4\sin (x) + 3 = 0\]
\[ 2\sin(x)\cos(x) + 4\sin (x) + 3 = 0\]
\[2\sin (x)\left(\cos(x)+2\right)+3=0\]
\[2\sin (x)\left(\cos(x)+2\right)=-3\]
\[\cos(x)+2=\frac{-3}2\csc(x)\]
i dont know

- anonymous

@satellite73 @saadi

- Hero

I don't understand why @UnkleRhaukus can't finish it

- anonymous

Can u Finish @Hero i am also stuck..there

- anonymous

@jim_thompson5910

- Mimi_x3

hmm
http://www.wolframalpha.com/input/?i=sin2x+%2B+4sin+x+%2B+3+%3D+0
i dont think its possible..

- anonymous

i think..tthere is mistake in question..))))

- anonymous

@Lulu212 can u check ur question...))

- Mimi_x3

I think it should be \(sin^2x + 4sinx +3 =0\)

- Mimi_x3

then \(u=sinx\) more reasonable

- anonymous

then it is a quad eq

- anonymous

we can find sinx

- Hero

I'm inclined to agree with @Mimi_x3

- anonymous

this question can be solved ............why u guys are worrying????????

- anonymous

it cant be solved just endless ways of writing the equation

- anonymous

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.

- UnkleRhaukus

\[\sin^2x+4\sin x+3=0\]
\[u^2-4u+3=0\]
\[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\]
\[u=2\pm\frac{\sqrt{4}}{2}\]\[u=2\pm1\]\[u=1,3\]
\[(1)^2+4(1)+3=0,\qquad\qquad 3^2-4(3)+3=0\]\[8=0,\qquad\qquad\qquad 3^2-4(3)+3=0\]\[\text{false}\qquad\qquad\qquad 9-12+3=0\]\[\qquad\qquad\qquad\qquad0=0\]\[\qquad\qquad\qquad\text{true}\]

- anonymous

the equation toolbar in this thing sucks hairy ______. how do you write it so neatly like that?

- UnkleRhaukus

so \(u=\sin(x)=3\)
\[x=\arcsin (3)\]
... this question is very confusing
\[\text{\[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\]}\]

- UnkleRhaukus

oh damn
my solution is way off
\[\sin^2(\arcsin 3) + 4\sin (\arcsin 3) + 3 = 24â‰ 0\]

- UnkleRhaukus

oh
\(u^2+4u+3=0\) NOT \(u^2-4u+3=0\)

- anonymous

but this wasnt the problem that was stated lol

- UnkleRhaukus

\[u_{1,2}=-2\pm1=-3,-1\]

- UnkleRhaukus

\[\sin x=u=-3,-1\]
\[x_{1,2}=\arcsin (-1),\quad \arcsin(-3)\]
\[x_1=-\frac \pi2\qquad, x_2=\text{some transcendental nonsense} \]
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 \[\sin^2x + 4\sin x + 3 = 0\]

- UnkleRhaukus

ta-da

- Hero

People who post questions need to learn how to use the equation editors to avoid such confusion.

- anonymous

couldn't agree more @Hero

- UnkleRhaukus

i think someone needs to write a program that can interpet questions and convert them to \(\LaTeX\) automagically

- anonymous

@lulu212 i am disappoint at this equation

- Hero

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 \(\sin^2(x)\)?

- anonymous

beyond OS' capacity :P
we can ask Wolfram though. That dude is probably the smartest programmer I've seen alive

- Hero

That's because he was reincarnated Einstein

- anonymous

hahaha... I think he was already born when Einstein was still alive

- Hero

Well, Einstein's lost twin

- anonymous

this question has like 40 responses all because of a typing error

- Hero

We're still assuming that it is a typing error

- anonymous

it is

- Hero

There's no such thing as typing errors. Only human errors.

- anonymous

the person who asked for help did not even say anything after the question was posted (TYPOGRAPHICAL ERROR)

- UnkleRhaukus

can't we all be einsteins lost twin?

- anonymous

I know that I can't... I have a roach's brain compare to his

- anonymous

same difference hero lol

- UnkleRhaukus

twins dont necessarily have the same charistics

- anonymous

oh that's right... I'd be the opposite of his intellect then

- UnkleRhaukus

*
characteristic

- Hero

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.

- UnkleRhaukus

oh my.

- anonymous

where's David Hume?

Looking for something else?

Not the answer you are looking for? Search for more explanations.