Trig general solution

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August 26th 2012, 09:17 AM
srirahulan

Trig general solution

I have a risk at these two Trigonometric equation to find out the general solution.

$1.\ sin2x-cos2x-sinx+cosx=0\\ \ 2.\ 4cos3x-sin2xcosx=0$
August 26th 2012, 10:14 AM
Plato

Re: Trig general solution

Quote:

Originally Posted by srirahulan

I have a risk at these two Trigonometric equation to find out the general solution.
$1.\ sin2x-cos2x-sinx+cosx=0\\ \ 2.\ 4cos3x-sin2xcosx=0$

Have a look at this.
August 26th 2012, 01:30 PM
MaxJasper

Re: Trig general solution

Principal general solutions in the complex plane found as follows:

Substitute sin() & cos() functions with their equivalents in complex plane:

$z=x+\text{i y}$

$2 {Cos}({n$\theta $})=\left(z^n+\frac{1}{z^n}\right)$

$2 i {Sin}({n$\theta $})=\left(z^n-\frac{1}{z^n}\right)$

to obtain:

$\frac{1}{2 i}\left(z^2-\frac{1}{z^2}\right)-\frac{1}{2}\left(z^2+\frac{1}{z^2}\right)-\frac{1}{2 i}\left(z-\frac{1}{z}\right)+\frac{1}{2}\left(z+\frac{1}{z} \right) =0$

Solve to obtain the 4 principal roots:

$\{z\to i\},\{z\to 1\},\left\{z\to \frac{1}{2} \left(-i-\sqrt{3}\right)\right\},\left\{z\to \frac{1}{2} \left(-i+\sqrt{3}\right)\right\}$

Similarly, the 2nd equation will become:

$2\left(z^3+\frac{1}{z^3}\right)-\frac{1}{2i}\left(z^2-\frac{1}{z^2}\right)*\frac{1}{2}\left(z+\frac{1}{z }\right)=0$

with 6 principal roots:

$z=-i, z=i, z=-\frac{2+i}{\sqrt{5}}, z=\frac{2+i}{\sqrt{5}}, z=-\frac{3-2 i}{\sqrt{13}}, z=\frac{3-2 i}{\sqrt{13}}$
August 26th 2012, 06:08 PM
srirahulan

Re: Trig general solution

I cant understand your solution please give me the way to solve the equation and find the general solution.
August 26th 2012, 06:29 PM
MaxJasper

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Re: Trig general solution

http://mathhelpforum.com/attachment....1&d=1346034520
August 26th 2012, 07:47 PM
Vlasev

Re: Trig general solution

I'll try to explain somewhat what is going on. On the complex plane we have that

$\cos(nx) = \frac{e^{inx}+e^{-inx}}{2},\quad \sin(nx) = \frac{e^{inx}-e^{-inx}}{2i}$

If you let $z = e^{ix}$ we get this representation

$\cos(nx) = \frac{z^n+z^{-n}}{2},\quad \sin(nx) = \frac{z^n-z^{-n}}{2i}$

Next, we substitute these in your equations. The first one is

$\frac{z^2-z^{-2}}{2i}-\frac{z^2+z^{-2}}{2}-\frac{z-z^{-1}}{2i}+\frac{z^1+z^{-1}}{2} = 0$

After we multiply everything by $2iz^2$ and expand, the equation becomes

$-(1+i) + (1+i)z -(1-i) z^3 + (1-i)z^4 = 0$

The second equation becomes

$4\frac{z^3+z^{-3}}{2}- \frac{z^2-z^{-2}}{2i}\frac{z+z^{-1}}{2}= 0$

After we multiply everything by $4iz^3$ and expand, the equation becomes

$(1 + 8 i) + z^2 - z^4 - (1-8i) z^6 = 0$

Solving each of these equations gives you the roots that MaxJasper wrote (I used a math program to find them for me). These equations need to hold simultaneously, so the only roots you accept are ones that solve both equations. If you inspect them, you will see that only $z = i$ is common to both. Now, we have that $e^{ix} = i$ , so you are looking for solutions of this new equation

$e^{ix} = i$

Knowing a bit about the complex exponential, you will immediately see that the solutions to this new equation are

$x = \frac{\pi}{2} + 2\pi n, \quad n = 0, \pm 1, \pm 2, \dots$

As a final step you need check which of these solves the original equations, if any. This is indeed the case.