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Equation ${\displaystyle x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}}$
Using ${\displaystyle x=2\cos \left(\theta \right)}$ we get ${\displaystyle 2\cos \left(\theta \right)=\sqrt{2+\sqrt{2-\sqrt{2+2\cos \theta }}}}$
${\displaystyle 2\cos \theta =\sqrt{2+\sqrt{2-2\cos \frac{\theta }{2}}}}$
${\displaystyle 2\cos \theta =\sqrt{2+2\sin \frac{\theta }{4}}}$
${\displaystyle 4{\cos }^2\theta =2+2\sin \frac{\theta }{4}}$
After this step I am not able to solve it

Can you solve ${\displaystyle \sin x=\sin \left(\frac{x}{2}\right)}$ without a sum/double/half angle identity?

Prove ${\displaystyle \tan \left({\sin }^{-1}t\right)=\frac{t}{\sqrt{1-t^2}}}$

Limit of ${\displaystyle \sin \left(\sqrt{x+2}\right)-\sin \left(\sqrt{x+4}\right)\text{ as }x\to \mathrm{\infty }}$

How to prove ${\displaystyle \sin (x-a)+\sin (x+a)=2\sin \left(x\right)\cos \left(a\right)}$?

${\displaystyle \cos 5=\sqrt{1-a^2}}$ which statement is true?
1) ${\displaystyle \sin 20=4a\sqrt{1-a^2}\sqrt{1-4a^2}}$
2) ${\displaystyle \tan 50=\frac{1-a}{1+a}}$
3) ${\displaystyle \tan 10=\frac{2}{a\sqrt{1-a^2}}}$
4) ${\displaystyle \cos 65=\frac{12}{\sqrt{1-a^2}-\sqrt{3}a}}$

If ${\displaystyle \frac{a+b}{\csc x}=\frac{a-b}{\cot x}}$ show that ${\displaystyle \csc x\cot x=\frac{a^2-b^2}{4ab}}$

Find minimum ${\displaystyle f\left(x\right)=\sin \left(x\right)+\cos \left(x\right)+\sin \left(2x\right)+1}$ for ${\displaystyle x\in [0,2\pi ]}$

Verify ${\displaystyle \cos \left(x\right)=\frac{1-t^2}{1+t^2}}$ with ${\displaystyle t=\tan \left(\frac{x}{2}\right)}$

Prove ${\displaystyle \cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}$

How to show f′(0) exists for ${\displaystyle f\left(x\right)=e^{\frac{-1}{x^2}}\sin \left(\frac{1}{x}\right)}$ for ${\displaystyle x\ne 0}$ and f(0)=0?

Simplify ${\displaystyle \frac{1}{1-z}}$, given ${\displaystyle z=\cos \vartheta +i\sin \vartheta }$
Im

How to evaluate limit without LHospitals rule, but using ${\displaystyle \underset{x\to 0}{lim}\frac{\sin x}{x}=1}$
I am having hard time evaluating this limit:
${\displaystyle \underset{x\to 0}{lim}\frac{\cos 3x-1}{\cos 2x-1}}$
I know, that this fact can help
${\displaystyle \underset{x\to 0}{lim}\frac{\sin x}{x}=1}$
But how? We can transform the first expression to
${\displaystyle \underset{x\to 0}{lim}\frac{\cos 3x-1}{-2{\sin }^{2}x}}$
but what should I do next? I cannot use LHopitals rule for solving this.

Finding the range of ${\displaystyle f\left(x\right)=\frac{{\sin }^{2}x+\sin x-1}{{\sin }^{2}x-\sin x+2}}$
I am attempting to find the range of the function ${\displaystyle f\to \mathbb{R}\to \mathbb{R}}$ which is given by the expression as follows.
${\displaystyle f\left(x\right)=\frac{{\sin }^{2}x+\sin x-1}{{\sin }^{2}x-\sin x+2}}$
This is rational function in ${\displaystyle \sin x}$, so it will be periodic. Usually when there is a rational function with quadratic expressions, differentiation can be used or, x is rewritten in terms of y, so that gives.
${\displaystyle \sin x=\frac{(y+1)\pm \sqrt{10-2y-3y^2}}{2(y-1)}\in [-1,1]}$
This is a tedious task and so is using differentiation to find minima and maxima. Any hints to go about this problem using a different technique are appreciated.

how is this solved? ${\displaystyle \underset{x\to 0}{lim}\frac{1}{x^2}(\left({\tan (x+\frac{\pi }{4})}^{\frac{1}{x}}-e^2\right)}$

Finding the ${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{\sin x\ln (1+x)}}$ using Taylors

Evaluate ${\displaystyle \int {\cos }^2\left(x\right){\tan }^3\left(x\right)dx}$ using trigonometric substitution

Find an equation for the plane containing the two (parallel) lines
${\displaystyle v_1=(0,1,-2)+t(2,3,-1)}$
${\displaystyle v_2=(2,-1,0)+t(2,3,-1)}$

Assume that T is a linear transformation. Find the standard matrix of T. ${\displaystyle T:R^2\to R^4,T\left(e_1\right)=(3,1,3,1)}$ and ${\displaystyle T\left(e_2\right)=(-5,2,0,0)}$, where ${\displaystyle e_1=(1,0)}$ and ${\displaystyle e_2=90,1)}$

Find the vector by using properties of cross products.
${\displaystyle k\times (i-2j)}$