A trig. question

**smmxwell** · June 7th 2009, 06:56 PM

For range $0`<=A<=360`$ , solve:
$3cos^2 2A=1+sin A$

Much help is appreciated. Thanks

**mr fantastic** · June 7th 2009, 07:05 PM

Originally Posted by smmxwell

For range $0`<=A<=360`$ , solve:
$3cos^2 2A=1+sin A$

Much help is appreciated. Thanks

Substitute $\cos^2 A = 1 - \sin^2 A$ and re-arrange into an equation that looks something like the one in this thread: http://www.mathhelpforum.com/math-he...sin-x-3-a.html

That thread tells you what to do next.

**smmxwell** · June 7th 2009, 07:54 PM

Yeah. I know about that. Just that it's sin^2 2A. Even if i use the double angle formula, I still can't get anywhere. I know about factoring a q.e. but I just can't rearrange it.

**mr fantastic** · June 7th 2009, 08:01 PM

Originally Posted by smmxwell

Yeah. I know about that. Just that it's sin^2 2A. Even if i use the double angle formula, I still can't get anywhere. I know about factoring a q.e. but I just can't rearrange it.

My mistake. I misread the question. Are you expected to get exact solutions?

**smmxwell** · June 7th 2009, 08:35 PM

Yeah. The answer is A= 30, 150, 221 deg. 49', 318 deg. 11'.
I did my own steps and arrived at
$3cos^2 2A=(cos^2 A)/(1-sin A)$

Even though $cos^2 2A$ is definitely not equal to $cos^2 A$ , still, if I eliminate them, I manage to find $sin A= -2/3$ , which would yield the last two readings. So, any help?

btw, I just came back from a long break, so I can't really think straight now. I found another question that needs help too.

Solve $tan x + 2 sin x=0$ for range=1 circle.
How do you go further from $sin x +sin 2x= 0$

**Amer** · June 7th 2009, 09:18 PM

Originally Posted by smmxwell

Yeah. The answer is A= 30, 150, 221 deg. 49', 318 deg. 11'.
I did my own steps and arrived at
$3cos^2 2A=(cos^2 A)/(1-sin A)$

Even though $cos^2 2A$ is definitely not equal to $cos^2 A$ , still, if I eliminate them, I manage to find $sin A= -2/3$ , which would yield the last two readings. So, any help?

btw, I just came back from a long break, so I can't really think straight now. I found another question that needs help too.

Solve $tan x + 2 sin x=0$ for range=1 circle.
How do you go further from $sin x +sin 2x= 0$

I will solve a question is near to your see how I solve it

$\sqrt{3}tan x =\frac{sinx }{cosx}$

$\sqrt{3} \frac{sinx}{cosx} + 2sinx =0$

$\sqrt{3}sinx + 2sinx cos x =0$

$sinx(\sqrt{3}+2cosx)=0$

$sinx=0......x=n\pi$

$\sqrt{3}+2cosx=0......cosx=-\frac{\sqrt{3}}{2}......x=....$

your's is like this

**smmxwell** · June 7th 2009, 09:26 PM

I'm sorry but I don''t really follow. You mean you're giving me a similar question which has the same steps? Either way, I don't see how you get $ \sqrt{3} \frac{sinx}{cosx} + 2sinx =0 $ after $ \sqrt{3}tan x =\frac{sinx }{cosx} $

the 2 question I posed weren't related. they are different questions which I both don't know how to do.

**Amer** · June 7th 2009, 09:57 PM

Ok I do a mistake in my solution

${\color{red}\sqrt{3}tan x =\frac{sinx }{cosx}}$ I forgot to multiply the answer with $\sqrt{3}$

${\color{blue}\sqrt{3}tan x =\sqrt{3}\left(\frac{sinx }{cosx}\right)}$ the correct one

I will solve two of them Step by Step so look

$\sqrt{3}tan x + 2 sin x =0$

$\sqrt {3} \left(\frac{sinx}{cosx}\right) + 2 sinx =0$

$\sqrt{3} sinx + 2 sinx cosx =0$

$sinx ( \sqrt{3} + 2 cos x ) =0$

sinx =0 so x=n(pi)
cosx = -square(3)/2

your's

$tanx + 2 sinx =0$

$\frac{sinx}{cosx} + 2sinx =0$

$sinx + 2 sinx cosx =0$

$sinx (1+ 2cosx ) =0$

$sinx =0 ......$

$1+2cosx =0.....cosx=\frac{-1}{2}$

it is ok or not

**smmxwell** · June 7th 2009, 10:11 PM

Yes. I get the 2nd one. But I still don't understand the first one;

$ 3cos^2 2A=1+sin A $

**Amer** · June 7th 2009, 10:27 PM

Originally Posted by smmxwell

Yes. I get the 2nd one. But I still don't understand the first one;

$ 3cos^2 2A=1+sin A $

I will not solve it completely I will give you hints just

$3cos^2(2A)=1+sinA$

you should know that

$cos(2A)=1-2sin^2(A)$ find the

$(1-2sin^2(A))^{2}$

then plug it in your equation you will have an equation

you will have

$3(1-4sin^2(A)+4sin^4(A))=1+sin(A)$

$3-12sin^2(A)+12sin^4(A)=1+sin(A)$

rearrange it and let t=sin(A) you will have

$12t^4-12t^2-t+2=0$ this for you

**Grandad** · June 7th 2009, 10:31 PM

Hello smmxwell

Originally Posted by smmxwell

For range $0`<=A<=360`$ , solve:
$3cos^2 2A=1+sin A$

Much help is appreciated. Thanks

If the answers you gave are correct then there's a mistake in the question. None of them satisfies $3\cos^2 2A=1+\sin A$ . (Try it and see!) However:

$3\cos2A = 1+\sin A$

$\Rightarrow 3(1-2\sin^2A)=1+\sin A$

$\Rightarrow 6\sin^2A +\sin A -2 = 0$

$\Rightarrow (2\sin A -1)(3\sin A +2)=0$

$\sin A = 0.5, -0.6667$

$A = 30^o, 150^o, 211.81^o, 318.19^o$

Grandad

**smmxwell** · June 7th 2009, 10:44 PM

Yeah thats the anwser. I was trying to type in degrees and minutes.
$ \sin A = 30^o, 150^o, 211.81^o, 318.19^o $

@Amer. I don't think quartic eq. is in my syllabus. I'm sure there's go to be a more elegant and simpler way?

@ Grandad,

The steps are for
$ 3\cos2A = 1+\sin A $

Whereas, the question wants $ 3\cos^2 2A=1+\sin A $

your answer is correct though..

**Amer** · June 7th 2009, 10:52 PM

Originally Posted by smmxwell

Yeah thats the anwser. I was trying to type in degrees and minutes.
$ \sin A = 30^o, 150^o, 211.81^o, 318.19^o $

@Amer. I don't think quartic eq. is in my syllabus. I'm sure there's go to be a more elegant and simpler way?

@ Grandad,

The steps are for
$ 3\cos2A = 1+\sin A $

Whereas, the question wants $ 3\cos^2 2A=1+\sin A $

your answer is correct though..

so you gave me a wrong equation that is not my fault I solve

$3cos^2(2A)=1+sinA$ since you gave me it

but you did not gave me

$3cos(2A)=1+sinA$

in math usually there is more than one way to solve questions that refer to the solver of the problem

**smmxwell** · June 7th 2009, 10:56 PM

No. The question is $ 3cos^2(2A)=1+sinA $ and only that.

Ok. Now I understand Grandad's post. The question is wrong. Alright. That means the question intended should be $ 3\cos2A = 1+\sin A $ . Thanks a lot guys. Lol. I just spent the whole day trying to solve a wrong question.

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