Trig Equation Help

[Ximei]

Ok , so this equation has gave me a real headache.

\(\displaystyle 2cos^3x+sinx-3sin^2xcosx=0\)

I can't get anywhere near the solution.
I have already tried a multiplication with cos x and that got me nowhere :roll:
And I also tried to write
\(\displaystyle 3sin^2xcosx\)
as
\(\displaystyle 2sin^xcosx - sin^2xcosx\)

and got to \(\displaystyle 2cosxcos2x+sinxsin2x=0\) but I don't know if
it is right or how to continue from here.
Dose anybody have a clue how to resolve this?
Any help will be greatly appreciate .

 

[arthur ohlsten]

let us see if the equation is true for a given point;

2cos^3x + sin x -3 sin^2 x cosx =?0 let x = 90 degrees
0+1+0=?0
the equation is not true at all points, so we must find the point where the relationship holds.
factor out cos x
cos x [2 cos^2 x - 3 sin^2x] =-sin x
[2[1-sin^2x]- 3 sin^2x] = -sinx / cos x
[2-5 sin^2 x]= - tan x
5 sin^2 x -2 = tanx

I don't believe this can be solved in closed form. [ I could be wrong but I don't think I can.] If you plot tan x and [5 sin^2x-2]
on the same graph , the curves cross at a value just less than pi. By various methods you can determine the point to whatevewr accuracy is required

Sorry I coulsn't be of more help
Arthur

 

[mmm4444bot]

?

I'm wondering whether the original equation picked up a typographical error somewhere along the way.

(For examle, if the sine term were to be cubed, the equation would be much easier to solve.)

If we graph the equation as given, we see a period of 2? and six solutions per period. At first glance, it might seem like four solutions, instead of six. We need closer inspections around x = 1 and x = 4.


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Decimal approximations for x in the interval [0, 2?) are:

1.01722

1.10715

2.58802

4.15881

4.24874

5.72961

?

 

[mmm4444bot]

?

I just asked MVR5 to solve this equation, and it found closed forms for six solutions in the interval [-?, ?).

Two of the solutions look straightforward, but the other four do not!

\(\displaystyle x = arctan(2)\)

or

\(\displaystyle x = arctan(2 - \pi)\)

I had to look through some routines, to find out how the remaining four solutions came about. Interestingly, these Real solutions were calculated using some Complex numbers with imaginary parts. MVR5 expressed the remaining four solutions using the following formula and values for n and m.

\(\displaystyle x \ = \ -i \cdot ln \left(\frac{m + i \cdot n}{\sqrt{m^2 + n^2}} \right)\)

\(\displaystyle n \ = \ -\frac{1}{10}\sqrt{50 + 10\sqrt{5}} \left( -\frac{1}{2} + \frac{\sqrt{5}}{2} \right) \quad \text{and} \quad m \ = \ \frac{1}{10}\sqrt{50 + 10\sqrt{5}}\)

\(\displaystyle n \ = \ \frac{1}{10}\sqrt{50 + 10\sqrt{5}} \left( -\frac{1}{2} + \frac{\sqrt{5}}{2} \right) \quad \text{and} \quad m \ = \ -\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\)

\(\displaystyle n \ = \ -\frac{1}{10}\sqrt{50 + 10\sqrt{5}} \left( -\frac{1}{2} - \frac{\sqrt{5}}{2} \right) \quad \text{and} \quad m \ = \ \frac{1}{10}\sqrt{50 + 10\sqrt{5}}\)

\(\displaystyle n \ = \ \frac{1}{10}\sqrt{50 + 10\sqrt{5}} \left( -\frac{1}{2} - \frac{\sqrt{5}}{2} \right) \quad \text{and} \quad m \ = \ -\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\)


?
All of this gives me even more of a sense that the intended equation somehow got jumbled, unless, of course, the actual assignment is to find solutions using graphing technology.


?

 

[Ximei]

Thank you for your help
And looking through this I think I have to agree with mmm4444bot,
I guess that the equation had some error in it ,
I though about the sine term being cubed ,but
I wasn't sure if I was overlooking a possibility to solve the equation,
So I had to ask for help.
Thanks again