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

If $H(\mathrm{\theta <!--\; \theta \; -->})=\mathrm{\theta <!--\; \theta \; -->}\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$, find $H^{\prime }(\mathrm{\theta <!--\; \theta \; -->})$ and $H^{″}(\mathrm{\theta <!--\; \theta \; -->})$.

Is ${\tan }^{2}x=se{c}^{2}x-1$ an identity?

The solutions of the equation $\sin <!--\; \; -->x+\sin <!--\; \; -->3x=\frac{8}{3\sqrt{3}}$ are?

Trigonometric equation with sine graph
If ${\sin }^{-<!--\; -\; -->1(1/2)}={30}^{\circ <!--\; \circ \; -->}$ ,solve the equation I've never seen anything like this and I can't find anything in my textbook so an explanation would be appreciated

Сonvert trigonometric equation to quadratic equation
I am struggling to know how to transform this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function.
Question:
${\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}/2$ + $\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ + $1$ = $0$
In must be transform in pure quadratic form:
$t^2+t+1=0$

Principal solution of given trigonometric equation
Find the principal solution of the following equation:
${\sin }^2<!--\; \; -->x+2{\tan }^2<!--\; \; -->x+\frac{4}{\sqrt{3}}\tan <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x+\frac{11}{12}=0$
I tried it by converting both $\sin <!--\; \; -->x$ and $\tan <!--\; \; -->x$ into half angles ( of $\tan <!--\; \; -->x/2$) but it is resulting in ugly equation. Could someone suggest a smarter approach without using half angles?

Solve this Trigonometric equation
I am not quite good at maths, so can you help me ?
$\tan <!--\; \; -->x+2\cot <!--\; \; -->x-<!--\; -\; -->3=0$

When can you divide both sides of an equation by a trigonometric function??
I was told to never divide both sides of an equation by a trigonometric function in case I accidentally eliminate an answer (or divide by $0$).
However, why do we divide both sides by $\cos <!--\; \; -->x$ for $\sin <!--\; \; -->x=\cos <!--\; \; -->x$. How do I know whether I can divide or not by a trigonometric function?
Thank you.

Help solving trigonometric equations
I have stumbled into this equation
$sin(x)sin(12)-<!--\; -\; -->sin(24)sin(228-<!--\; -\; -->x)=0$
all what I managed to do is to simplify it to
$sin(x)-<!--\; -\; -->2sin(228-<!--\; -\; -->x)cos12=0$
and all what I did after was useless
my question is how to solve this equation and what exactly should I learn to solve equations like these?

Find the arclength of the curve $r(t)=<-<!--\; -\; -->9\sin <!--\; \; -->t,2t,-<!--\; -\; -->9\cos <!--\; \; -->t>,-<!--\; -\; -->8\le <!--\; \le \; -->t\le <!--\; \le \; -->8$

Question in a trigonometric equation.
I tried it a lot but am not able to get this.Pls help in how should I think when solving this type of question and which side is better to try to simplify first (LHS or RHS).Please share the solution in that way.
$\frac{1-<!--\; -\; -->\sin <!--\; \; -->A}{1+\sin <!--\; \; -->A}=1+2\tan <!--\; \; -->A(\tan <!--\; \; -->A-<!--\; -\; -->\sec <!--\; \; -->A)$
One of the ways I tried but am not understand that how to simplify it in such a way that you get the RHS?

Trigonometric equation.
Please help in Solving the Trigonometric Equation:
${\cos }^2<!--\; \; -->x-<!--\; -\; -->{\sin }^2<!--\; \; -->x=\cos <!--\; \; -->3x$

For $y=-<!--\; -\; -->2\sin <!--\; \; -->(x-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$

Inverse trigonometric equation
${\tan }^{-<!--\; -\; -->1}<!--\; \; -->x+{\cot }^{-<!--\; -\; -->1}<!--\; \; -->x=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$
I think there have to be used formula:
${\displaystyle \tan <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->})=\frac{\tan <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}+\tan <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}}{1-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\tan <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}}}$ but I don't know how to apply it and yet I don't know whether it's true that
${\displaystyle {\cot }^{-<!--\; -\; -->1}<!--\; \; -->x=\frac{1}{{\tan }^{-<!--\; -\; -->1}<!--\; \; -->x}}$?

trigonometric equation using compound angle formula
If $\arcsin <!--\; \; -->(12/13)+\arccos <!--\; \; -->(4/5)=\arcsin <!--\; \; -->(k)$
Use the compound angle formula $\sin <!--\; \; -->(x+y)=\sin <!--\; \; -->(x)\cos <!--\; \; -->(y)+\cos <!--\; \; -->(x)\sin <!--\; \; -->(y)$ to find the exact value of k

Try to find the number of solutions of the equation
$\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\sin <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}$
in the interval $[-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->},\mathrm{\pi <!--\; \pi \; -->}]$

Confusion with trigonometric equation
$\begin{array}{rlr}6{\sin }^2<!--\; \; -->x& =1& 0\le <!--\; \le \; -->x\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}\\ {\sin }^2<!--\; \; -->x& =1/6\\ \sin <!--\; \; -->x& =1/\sqrt{6}\\ \sin <!--\; \; -->a& =1/\sqrt{6}\\ a& =0.420\end{array}$
so $x$ is $0.420$ and $2.721$
According to the key at the back of the book there are supposed to be 2 more answers; $3.56$ and $\mathrm{5.86.}$ Could someone explain where I went wrong?

Solving trigonometric equations: do I need to consider different cases in this example?
I'm trying to solve this set of trigonometric equations:
$\begin{array}{r}\begin{array}{rl}\tan <!--\; \; -->\left(\frac{\mathrm{\delta <!--\; \delta \; -->}+\mathrm{\beta <!--\; \beta \; -->}}{2}\right)& =c\cdot <!--\; \cdot \; -->\tan <!--\; \; -->\left(\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}\right)\\ b\cdot <!--\; \cdot \; -->\tan <!--\; \; -->\left(\frac{\mathrm{\delta <!--\; \delta \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}}{2}\right)& =a\\ \cos <!--\; \; -->\left(\frac{\mathrm{\delta <!--\; \delta \; -->}+\mathrm{\beta <!--\; \beta \; -->}}{2}\right)\cos <!--\; \; -->\left(\frac{\mathrm{\gamma <!--\; \gamma \; -->}}{2}\right)& =\cos <!--\; \; -->\left(\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}\right)\end{array}\end{array}$
Where a,b,c and $\mathrm{\theta <!--\; \theta \; -->}$ are some given (but random) real numbers (they could be $0$ and $\mathrm{\pi <!--\; \pi \; -->}$). The goal is to solve for the real values of $\mathrm{\gamma <!--\; \gamma \; -->},\mathrm{\beta <!--\; \beta \; -->}$ and $\mathrm{\delta <!--\; \delta \; -->}$. I'm trying to solve these equations by combining some variables. Let
$\begin{array}{r}M=\frac{\mathrm{\delta <!--\; \delta \; -->}+\mathrm{\beta <!--\; \beta \; -->}}{2},N=\frac{\mathrm{\delta <!--\; \delta \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}}{2},P=\frac{\mathrm{\gamma <!--\; \gamma \; -->}}{2}\end{array}$
then it's easy to solve for $M$ from equation $1$. However, to solve for $\mathrm{\delta <!--\; \delta \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$, we also need to figure out $N$. I'm worried that if I need to discuss the solutions by cases, like if $\mathrm{\theta <!--\; \theta \; -->}=\mathrm{\pi <!--\; \pi \; -->}$ (so that $\tan <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}$ is undefined) or if $a$ or $b$ is equal to $0$. Any suggestions on solving for this question? Is it possible to obtain a generalized solution? Thanks!!

trigonometric equation involving two variables
Show that the equation ${\sec }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{4xy}{(x+y{)}^2}$ is only possible when $x=y$
Can we write it this way :
${\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{(x+y{)}^2}{4xy}$
and
$-<!--\; -\; -->1\le <!--\; \le \; -->\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->1⟹<!--\; ⟹\; -->-<!--\; -\; -->1\le <!--\; \le \; -->\frac{(x+y{)}^2}{4xy}\le <!--\; \le \; -->1$
or $-<!--\; -\; -->4xy\le <!--\; \le \; -->(x+y{)}^2\le <!--\; \le \; -->4xy$ Please guide...

Solve a given trigonometric equation
Solve the following equation:
$\sin <!--\; \; -->x+\sin <!--\; \; -->(x+\frac{7\mathrm{\pi <!--\; \pi \; -->}}{24})=\frac{\sqrt{2-<!--\; -\; -->\sqrt{2}}}{2}+\frac{\sqrt{6}+\sqrt{2}}{4}$
So far, I found out that $\frac{\sqrt{2-<!--\; -\; -->\sqrt{2}}}{2}=\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}$ and $\frac{\sqrt{6}+\sqrt{2}}{4}=\sin <!--\; \; -->\frac{7\mathrm{\pi <!--\; \pi \; -->}}{12}$.
Thank you!