Demystify Trigonometric Equations and Identities with Step-by-Step Examples and Expert Guidance

Re-express the following as a rational number, that is a number of the form n/m, where m and n are integers:
$9\sinh <!--\; \; -->(\ln <!--\; \; -->12)$

Find the exact value of the expression.
$\tan <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}-<!--\; -\; -->\cos <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$

Find all solutions of the following equation
$2\sqrt{3}+\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\sqrt{3}$

Solve the trigonometric equation
Solve the equation
$\cos <!--\; \; -->x-<!--\; -\; -->2\cos <!--\; \; -->2x+3\cos <!--\; \; -->3x-<!--\; -\; -->4\cos <!--\; \; -->4x={\displaystyle \frac{1}{2}}.$
I tried, put $t=\cos <!--\; \; -->x$

I'm stuck at a relatively simple equation where I was simplifying like this:
${\cos }^2<!--\; \; -->(x)+2\sin <!--\; \; -->(x)\cos <!--\; \; -->(x)-<!--\; -\; -->{\sin }^2<!--\; \; -->(x)=1⟺<!--\; ⟺\; -->\sin <!--\; \; -->(2x)+\cos <!--\; \; -->(2x)=1$
How to I continue?

Solve: $\sqrt{3}\sin <!--\; \; -->x-<!--\; -\; -->\cos <!--\; \; -->x=2$, $-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}<x<2\mathrm{\pi <!--\; \pi \; -->}$
My attempt:
$\sqrt{3}\sin <!--\; \; -->x-<!--\; -\; -->\cos <!--\; \; -->x=2$
$\frac{\sqrt{3}}{2}\sin <!--\; \; -->x-<!--\; -\; -->\frac{1}{2}\cos <!--\; \; -->x=1$
$\sin <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}).\sin <!--\; \; -->x-<!--\; -\; -->\cos <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{3})\cos <!--\; \; -->x=1$
$\cos <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{3})\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}).\sin <!--\; \; -->x=-<!--\; -\; -->1$
$\cos <!--\; \; -->(x+\frac{\mathrm{\pi <!--\; \pi \; -->}}{3})=\cos <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}$
$x+\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}=2n\mathrm{\pi <!--\; \pi \; -->}\pm <!--\; \pm \; -->\mathrm{\pi <!--\; \pi \; -->},\text{where }n\text{ belongs to }\mathbb{Z}$
$x=2n\mathrm{\pi <!--\; \pi \; -->}\pm <!--\; \pm \; -->\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$
Now, if n=0,
$x=\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3},-<!--\; -\; -->\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3}$
If n=1,
$x=\frac{8\mathrm{\pi <!--\; \pi \; -->}}{3},\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3}$
If n=-1,
$x=-<!--\; -\; -->\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3},-<!--\; -\; -->\frac{10\mathrm{\pi <!--\; \pi \; -->}}{3}$
So, in the given interval, $x=\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3},\frac{-<!--\; -\; -->4\mathrm{\pi <!--\; \pi \; -->}}{3}$
Question:
When n=0, and when n=1, I get $\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3}$. Similarly, when n=0, and when n=−1, I get $-<!--\; -\; -->\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3}$. Why am I getting the same angle twice for different values of n?

Difference equation with trigonometric term
Trying to find the general solution to this homogeneous difference equation:
$y_k-<!--\; -\; -->2\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{y}_{k-<!--\; -\; -->1}+y_{k-<!--\; -\; -->2}=0.$
The characteristic equation is
${\mathrm{\lambda <!--\; \lambda \; -->}}^2-<!--\; -\; -->2\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\mathrm{\lambda <!--\; \lambda \; -->}+1=0.$
Not sure how to factor this, but tried
$(\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})(\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})=0$
but I am stuck as to how to get ${\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=1$ using trigonometric identities.
By using the quadratic formula I get a discriminant of
$4({\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->1)$
and I am stuck on how to simplify this to get the general solution.
Any help is appreciated. This is not for homework, but self study.

Solve for x: $2{\sin }^2<!--\; \; -->(x)-<!--\; -\; -->{\sin }^2<!--\; \; -->(2x)={\cos }^2<!--\; \; -->(2x)$

Simplifying a trigonometric equation for chart radius
I am making a neat new charting application, yet I am stumped by this equation:
$M={\displaystyle \frac{R}{\cos <!--\; \; -->\left(\frac{\theta }{2}\right)}}+(\tan <!--\; \; -->\left(\frac{\theta }{2}\right)\cdot <!--\; \cdot \; -->R)$
I need to solve for $R$, but for some reason I am throwing a blank. Any help?
Thanks!

Help with trigonometric equation $\tan <!--\; \; -->(x)+\cot <!--\; \; -->(2x)=1$
It is clear that the equation has a solution $x=\mathrm{\pi <!--\; \pi \; -->}/4$ but I can't show that this is the only solution. Any ideas?
P.S. I am trying to solve the equation without using formulas for $2a$ angle

Evaluating $\int <!--\; \int \; -->({\tan }^3<!--\; \; -->x+{\tan }^4<!--\; \; -->x)dx$ using substitution $t=\tan <!--\; \; -->x$

How to use parametric equation/trigonometric identity to show an ellipse?
I have the equation $16x^2+25y^2=400$, and the parametric equation $(x,y)=(5\cos <!--\; \; -->t,4\sin <!--\; \; -->t)$
If I plug in the parametric equation into the first equation, I end up with the trigonometric identity ${\cos }^2<!--\; \; -->t+{\sin }^2<!--\; \; -->t=1$. How does this identity show that my non-parametric equation, when graphed, will result in an ellipse?

Solve the equation
${\cot }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{3}{2}\csc <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$

Finding all constant solutions for the differential equation y'=sin(y)
For my homework of finding ALL the constant solutions of given differential equations, a trigonometric function popped up
$\frac{dy}{dx}=sin(y)$
Unfortunately i'm new to differential equations and I could only prove it true for y=0 but when I plotted it on Desmos I could see that it was true for every multiple of pi. I was wondering how I could prove that, and how to deal with finding constant solutions for differential equations with trigonometric fucntions in general.

Trigonometry: Solve equation for $\mathrm{\alpha <!--\; \alpha \; -->}$
I have the following trigonometric equation:
$2\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->45)\sin <!--\; \; -->(2\mathrm{\alpha <!--\; \alpha \; -->})=\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->}+45)\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})$
Is it possible to find $\mathrm{\alpha <!--\; \alpha \; -->}$?
Please also include each step in your solution.
EDIT: Sorry if I haven't mentioned- yes, it is a solution I reached to as a part of an assignment I was given (school). All I wish to know if I can pull $\mathrm{\alpha <!--\; \alpha \; -->}$ from what I found.
Thanks in advance.

Find all $n\in <!--\; \in \; -->\mathbb{Z}$ such that $k=\frac{1+4n}{5},(k\in <!--\; \in \; -->\mathbb{Z})$
My question is rather general but I got stuck in that issue after trying to solve a trigonometric equation.
After simplifying I got this:
$\sin <!--\; \; -->\left(\frac{5x}{4}\right)+\cos <!--\; \; -->x=2$
which is true if and only if
$\begin{array}{r}\{\begin{array}{ll}\sin <!--\; \; -->\left(\frac{5x}{4}\right)& =1\\ \cos <!--\; \; -->x& =1\end{array}\end{array}$
Hence
$\begin{array}{rl}\cos <!--\; \; -->x& =1\\ ⟹<!--\; ⟹\; -->x& =2\mathrm{\pi <!--\; \pi \; -->}k,(k\in <!--\; \in \; -->\mathbb{Z})\end{array}$
and
$\begin{array}{rl}\sin <!--\; \; -->\left(\frac{5x}{4}\right)& =1\\ ⟹<!--\; ⟹\; -->\frac{5x}{4}& =\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+2\mathrm{\pi <!--\; \pi \; -->}n\\ & =\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}+\frac{8\mathrm{\pi <!--\; \pi \; -->}n}{5},(n\in <!--\; \in \; -->\mathbb{Z})\end{array}$
Thus the solutions are
$\begin{array}{rl}2\mathrm{\pi <!--\; \pi \; -->}k& =\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}+\frac{8\mathrm{\pi <!--\; \pi \; -->}n}{5}\\ k& =\frac{1+4n}{5},(k\in <!--\; \in \; -->\mathbb{Z})\end{array}$
Here I'm stuck. Clearly the constraint of $k\in <!--\; \in \; -->\mathbb{Z}$ is not true for all n's. n has to be made up of some number m that makes the numerator divisible by 5. The answer that I'm given says that $n=5m+1$, which makes sense but I don't know how to get there. And what if, just to say, ${\displaystyle k=\frac{8+13n}{7}}$?
Thanks!!

Solving simultaneous trigonometric equations
Question: Solve the following equations for $x$ and $y$
$\sin <!--\; \; -->(y)=\sin <!--\; \; -->(x-<!--\; -\; -->y)$
$\cos <!--\; \; -->(x)=\sin <!--\; \; -->(x-<!--\; -\; -->y)$
where $x,y\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$
My attempt:
The beginning of the solution seems to have something to do with
$(\mathrm{i})y=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+x\vee <!--\; \vee \; -->(\mathrm{i}\mathrm{i})y=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->x$
However I cannot understand how this conclusion was made.

Find the exact solution of the equation.
$3{\cos }^{-<!--\; -\; -->1}<!--\; \; -->(4x)=2\mathrm{\pi <!--\; \pi \; -->}$

How do you solve the trigonometric equation $\sin <!--\; \; -->(x)+x=9$?
How do you solve the trigonometric equation $\sin <!--\; \; -->(x)+x=9$? More generally, how do you solve equations with both trigs and 'x's without graphing? And maybe I only want real number answers.
Suggestions and edit: If you use the Taylor Series, you will basically end up with at least $6^{th}$ degree equations, if you want to make it close. Equation Solvers Are NOT Allowed. That will kind of limit you to this:(assuming you make careless mistakes, just like me, so you don't want cubic equations)
$x+x=9.$
This yields $x=4.5$, which is DEFINITELY a bad estimation.
Thanks for helping me! It's just that I don't know what Newton-Ralphson and iterations are. Can you explain it to me or give me a link?

How to solve the trigonometric equation $\sin <!--\; \; -->x+\cos <!--\; \; -->x=\sin <!--\; \; -->2x+\cos <!--\; \; -->2x$?
My attempt:
$\sin <!--\; \; -->x+\cos <!--\; \; -->x=\sin <!--\; \; -->2x+\cos <!--\; \; -->2x$
$⟹<!--\; ⟹\; -->\sin <!--\; \; -->x+\cos <!--\; \; -->x=2\sin <!--\; \; -->x\cos <!--\; \; -->x+{\cos }^2<!--\; \; -->x-<!--\; -\; -->{\sin }^2<!--\; \; -->x$
$⟹<!--\; ⟹\; -->\sin <!--\; \; -->x+\cos <!--\; \; -->x=2\sin <!--\; \; -->x\cos <!--\; \; -->x+{\cos }^2<!--\; \; -->x-<!--\; -\; -->(1-<!--\; -\; -->{\cos }^2<!--\; \; -->x)$
$⟹<!--\; ⟹\; -->\sin <!--\; \; -->x+\cos <!--\; \; -->x=2\sin <!--\; \; -->x\cos <!--\; \; -->x+2{\cos }^2<!--\; \; -->x-<!--\; -\; -->1$
$⟹<!--\; ⟹\; -->\sin <!--\; \; -->x-<!--\; -\; -->2\sin <!--\; \; -->x\cos <!--\; \; -->x+\cos <!--\; \; -->x-<!--\; -\; -->2{\cos }^2<!--\; \; -->x=-<!--\; -\; -->1$
$⟹<!--\; ⟹\; -->\sin <!--\; \; -->x(1-<!--\; -\; -->2\cos <!--\; \; -->x)+\cos <!--\; \; -->x(1-<!--\; -\; -->2\cos <!--\; \; -->x)=-<!--\; -\; -->1$
$⟹<!--\; ⟹\; -->(1-<!--\; -\; -->2\cos <!--\; \; -->x)(\sin <!--\; \; -->x+\cos <!--\; \; -->x)=-<!--\; -\; -->1$
$⟹<!--\; ⟹\; -->(1-<!--\; -\; -->2\cos <!--\; \; -->x)=-<!--\; -\; -->1$ or $(\sin <!--\; \; -->x+\cos <!--\; \; -->x)=-<!--\; -\; -->1$
$⟹<!--\; ⟹\; -->x=2n\mathrm{\pi <!--\; \pi \; -->}$ or $\sin <!--\; \; -->2x=0$
$⟹<!--\; ⟹\; -->x=2n\mathrm{\pi <!--\; \pi \; -->}$ or $2x=n\mathrm{\pi <!--\; \pi \; -->}$
$\therefore <!--\; \therefore \; -->x=2n\mathrm{\pi <!--\; \pi \; -->}$ or $x=\frac{n\mathrm{\pi <!--\; \pi \; -->}}{2}$
But the answers given in my book are $x=2n\mathrm{\pi <!--\; \pi \; -->}$ and $x=\frac{(4n+1)\mathrm{\pi <!--\; \pi \; -->}}{6}$. Where have I gone wrong? Please help.