# Thread: A trig. question

1. ## A trig. question

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

Much help is appreciated. Thanks

2. Originally Posted by smmxwell
For range $\displaystyle 0<=A<=360$, solve:
$\displaystyle 3cos^2 2A=1+sin A$

Much help is appreciated. Thanks
Substitute $\displaystyle \cos^2 A = 1 - \sin^2 A$ 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.

3. 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.

4. 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?

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

Even though $\displaystyle cos^2 2A$ is definitely not equal to $\displaystyle cos^2 A$, still, if I eliminate them, I manage to find $\displaystyle sin A= -2/3$, 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 $\displaystyle tan x + 2 sin x=0$ for range=1 circle.
How do you go further from $\displaystyle sin x +sin 2x= 0$

6. 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
$\displaystyle 3cos^2 2A=(cos^2 A)/(1-sin A)$

Even though $\displaystyle cos^2 2A$ is definitely not equal to $\displaystyle cos^2 A$, still, if I eliminate them, I manage to find $\displaystyle sin A= -2/3$, 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 $\displaystyle tan x + 2 sin x=0$ for range=1 circle.
How do you go further from $\displaystyle sin x +sin 2x= 0$
I will solve a question is near to your see how I solve it

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

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

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

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

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

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

your's is like this

7. 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 $\displaystyle \sqrt{3} \frac{sinx}{cosx} + 2sinx =0$ after $\displaystyle \sqrt{3}tan x =\frac{sinx }{cosx}$

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

8. Ok I do a mistake in my solution

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

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

I will solve two of them Step by Step so look

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

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

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

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

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

your's

$\displaystyle tanx + 2 sinx =0$

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

$\displaystyle sinx + 2 sinx cosx =0$

$\displaystyle sinx (1+ 2cosx ) =0$

$\displaystyle sinx =0 ......$

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

it is ok or not

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

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

10. Originally Posted by smmxwell
Yes. I get the 2nd one. But I still don't understand the first one;

$\displaystyle 3cos^2 2A=1+sin A$
I will not solve it completely I will give you hints just

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

you should know that

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

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

then plug it in your equation you will have an equation

you will have

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

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

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

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

11. ## Trig equations

Hello smmxwell
Originally Posted by smmxwell
For range $\displaystyle 0<=A<=360$, solve:
$\displaystyle 3cos^2 2A=1+sin A$

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

$\displaystyle 3\cos2A = 1+\sin A$

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

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

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

$\displaystyle \sin A = 0.5, -0.6667$

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

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

@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?

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

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

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

@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?

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

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

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

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

but you did not gave me

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

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

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

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