Simplify and Solve Trig Equation

[Gib Z]

Homework Statement

Simplify and Solve the Following Trigonometric Equation

Homework Equations

\cos^2 x - 3\cos x - 2\sin x +2 =0

The Attempt at a Solution

I've changed the expression, but it doesn't seem any better...

I've got \frac{\cot x \cdot (\cos (x) -3)}{2}=1 - \csc x

But that doesn't seem to help me at all...Stuck badly.

 

[Moridin]

Solving equations that only contain sin x or cos x is easier than an equation that contains both.

 

[Gib Z]

Yes, I know how to solve those ones. Unfortunately this one didn't come this way...

 

[mjsd]

try using the fact that if t=\tan\frac{x}{2} then
\displaystyle{\cos x = \frac{1-t^2}{1+t^2}} and
\displaystyle{\sin x = \frac{2t}{1+t^2}}, solve for t, should be a quatic eqn and then you can solve for x.

NB: may not be the simplest way to do things, but at least in principle this will work. would like to see if there is a simpler way to do this...

 

[robphy]

Upon substituting \cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}}) and \sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}}), where i^2=-1, you end up with a quartic polynomial in one variable "e^{ix}", which generally has four solutions.

If the above method is not allowed or not what is expected, you probably have to rewrite the equation in terms of one trigonometric-function of x [using trig identities], as suggested by Moridin.

 

[Gib Z]

I have absolutely no idea how to do this using identities, just makes anything I do more messy and doesn't help. And I don't think this problem should require math from that level, but ill try it. Solving quartics are hard though...

 

[robphy]

I (hopefully correctly) typed in the equation to Maple and asked for solutions. Two of the four roots for x are real (and will probably arise simply).. the other two are complex and are not pretty (though trig function of these may be prettier).

 

[mjsd]

if you use my method, the quartic in t that you need to solve is actually very easy to do! try it! Hint: there are two real integer roots, the remaining complex roots can then be easily worked out using the quadratic equation. Finding x is then a matter of inverting the \tan.
NB: robphy's method is in essence the same; the only difference is that you introduce complex numbers from the very beginning.

 

[Gib Z]

Maybe there is a simpler way. I didn't actaully post my original question, but what I got it to correctly, here's the whole thing.

cos^3 x-3cos^2 x+ cos x = 2cos(x/2 + pi/4)sin(3x/2 - pi/4)
cos^3 x-3cos^2 x+ cos x= sin2x - cos x
cos^3 x-3cos^2 x+ 2cos x - 2sinxcosx=0
cos x(cos^2 - 3cos x +2 - 2sin x)=0

Now I'll do the first one, cos x=0, now I needed the 2nd part. Was there an easier way from the start?

Ill try using the t, looks good.