Trigonometric Equation

Very well Melody, in future, to anything that I post, I will attach the name Tiggsy.

Hi Quazars,

Yes I took a look but I have drawn a blank as well. 

The Wolfram Alpha solution is really messy.  I do not think that there is any neat solution.

https://www.wolframalpha.com/input/?i=8.8cos%5E2x-20sin(2x)%2Bc%3D0

Use the trig identity

\(\displaystyle \cos^{2}(x)=(1/2)(1+\cos(2x))\),

Therafter it's a routine equation of the type

a.cos(theta) - b.sin(theta) = c.

(Write it as Rcos(theta + alpha) = c and calculate R and alpha ).

Thanks guest :)

Do you understand our guest's answer Quazars?

I have just worked through guest's outline. :)

\(cos(2x)=cos^2x-sin^2x\\ cos(2x)=cos^2x-(1-cos^2x)\\ cos(2x)=2cos^2x-1\\ 0.5[cos(2x)+1]\\~\\ 8.8cos^2(x)-20sin(2x)+C=0\\ 8.8*0.5[cos(2x)+1]-20sin(2x)+C=0\\ 4.4cos(2x)+4.4-20sin(2x)=-C\\ 4.4cos(2x)-20sin(2x)=-(C+4.4)\\~\\ Let\;\;\\R[cos(2x+\theta)]=4.4cos(2x)-20sin(2x)\\ Rcos(2x)cos\theta-Rsin(2x)sin\theta=4.4cos(2x)-20sin(2x)\\ so\\ Rcos\theta=4.4 \qquad and \qquad Rsin\theta=20\\ R^2cos^2\theta=4.4^2 \qquad and \qquad R^2sin^2\theta=20^2\\ add\\ R^2cos^2\theta+R^2sin^2\theta=4.4^2+20^2\\ R^2(cos^2\theta+sin^2\theta)=419.36\\ R^2=419.36\\ r=\sqrt{419.36}\)

\(cos^2\theta=\frac{4.4^2}{419.36}\\ \theta\approx 77.59^\circ +180n \qquad n\in Z\\ \text{I should check each quadrant answer - the squaring i did probably invalidates some ot them.}\\~\\ R[cos(2x+\theta)]=-(4.4+c)\\ \sqrt{419.36}[cos(2x+77.59+180n^{\circ})]=-(4.4+c)\\ 2x+77.59+180n^{\circ}=acos\frac{-(4.4+c)}{\sqrt{419.36}}\\ 2x=acos\frac{+(4.4+c)}{\sqrt{419.36}}-77.59+(or-)180n^{\circ}\\ x=0.5\left[ \left(acos\frac{4.4+c}{\sqrt{419.36}}\right)-77.59+180n \right]\;\;degrees \qquad n\in Z\\ \)

\(\text{I haven't checked that this is true for all n, }\\\text{I am really just showing that there will be multiple answers :(}\)

*

Maybe I should just stick to the easiest answer 

\(\\ x=0.5\left[ \left(acos\frac{4.4+c}{\sqrt{419.36}}\right)-77.59\right]\;\;degrees\\ \)

This answer needs to be seriously checked.........  I've had enough for now

The multiple angle situation can be resolved by looking at the equations

\(\displaystyle R\cos(\theta) = 4.4 \text{ and }R\sin(\theta) = 20\).

If \(\displaystyle R\) is positive, then theta must be in the first quadrant.

It's more usual to find theta by dividing one by the other, \(\displaystyle \tan(\theta) = 20/4.4\) .

Yes, it was very helpful, I understood Guest's guidelines but I want to also thank you for providing the full steps and working out. 

Since the both of you took the time to help me out I hope I won't be an annoyance by asking how exactly you arrived at this part: 

\(Let\;\;\\R[cos(2x+\theta)]=4.4cos(2x)-20sin(2x)\)

A similar identity that I know of is A*sin(x) + B*cos(x) = C*sin(x+v) where tan(v) = b/a

Also, based on the answer you provided and plugging in C it ended up incorrect. 

In addition to the identity you state, there are two others.

Using your notation,

Asin (x) - Bcos (x) = Csin (x - v)

and

Acos (x) - Bsin (x) = Ccos (x + v ).

It was this last one that was chosen in order that the signs matched up.

Incidently, if the rhs of Melody's equation is known to be negative, (it depends on the value of c),  it might be better to switch the left and right hand sides. In that case we would use the identity for 

Asin (x) - Bcos(x).

Tiggsy