Question: The  number of real roots of the equation $$2\cos\left( \frac{x^2+x}{6}\right)=2^x+2^{-x}$$ is (i) $0$ $\qquad$ (ii) $1$   $\qquad$(iii)  $2$    $\qquad$ (iv)  $\infty.$      I have tried in the following way:       $$2\cos\left( \frac{x^2+x}{6}\right)=2^x+2^{-x}\implies \cos\left( \frac{x^2+x}{6}\right)=\frac{2^x+2^{-x}}{2}.$$ What can I do after this?
120774
Find the number of real roots of   $ \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}$  1) 0  2) 1  3) 2  4)  None of these  My guess is to approach it in graphical way. But equation seems little difficult.
93185
The graph of the function $f (x)$ is given.  How many different real roots does the $f (4-3x^2)=0$ equation have?  Here is possible solution:  $$\Bigg [{4-3x^2=-2 \\ 4-3x^2=4} \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$  Then, here is different solution:  $$f(x)=x^2-2x-8$$ $$f(4-3x^2)=0\Longrightarrow (4-3x^2)^2-2(4-3x^2)-8=0 \Longrightarrow 9x^4-18x^2=0 \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$   Problematic point. Here, I assumed the function is quadratic. But, obviously, this function may not be quadratic.  My questions:  Question $-1:$ Is the first solution completely correct?  Question $-2:$ Can we say that, the second solution is completely and definitely wrong? If so, is it possible to add something to this method and turn it into the right solution?
28157
So I'm trying to get this:    And I've calculated $z^4=16 \left( \cos (\frac{- \pi}{3})+ \sin ( \frac{- \pi}{3}) \right)$  So I'm trying to find the roots using the formula:  $ r^{1/n} = \left( \cos ( \frac { \theta + 2 \pi \cdot k}{n}) + i \cdot \sin ( \frac{ \theta + 2 \pi \cdot k}{n}) \right) $  But my result does not equal. Take a look:    This is for $k=0$.  What am I doing wrong?
114079
I am looking for direction in my question. The question is to prove, by long polynomial division, that: \begin{equation}  2^{2^{n+1}}-1\ |\ 2^{F_n-1}-1 \end{equation} I have tried a lot of things, like calling:\begin{equation}  t=2^{2^n} \end{equation} and trying long polynomial division right away, but none of this worked successfully.
26209
Next is what I have worked out to the moment.   $1$ and $-1$ are roots for all $n$.  $x \in \mathbb{Z}/n\mathbb{Z},\ $ $x^2\equiv1 \Leftrightarrow (x-1)(x+1)\equiv0 \Leftrightarrow \exists k \in \mathbb{Z}/n\mathbb{Z}: k(k+2)\equiv0 $. But how can it be applied to find other roots?
57873
Is there an easy way to prove the identity?     $$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$   While solving one , I am stuck, which looks obvious but without any feasible way to approach.   Few observations, not sure if it would help $$ \begin{align} \dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &amp;= \dfrac{2\pi}{7}\\\\ \dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &amp;= \pi - \dfrac{\pi}{7} \end{align} $$
124300
I am reading a proof that narrows down to the following statement:      It is easy to see that the number of solutions of $x^6+x=a$ in $\mathbb{F}_{2^m}$, where $a\in \mathbb{F}_{2^m}$ and $m\geq 3$ is odd, is same as number of solutions of $x^2+ax+1=0$ in $\mathbb{F}_{2^m}$.   Why is this true? They say that $x^7=1$ for $x \neq 0$, which I also don't see. Thank you!
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Problem Statement:- Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$   I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value. These are the things that I have tried to turn the given equation into a quadratic equation. $$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2$$ $$\text{OR}$$ $$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2$$
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$$ 0&lt;\theta&lt;\pi/2$$   and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$   then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$   I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$.   But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.  When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.  Even hints would work, as I would like to solve the question myself.
60388
$x$ and $y$ are real numbers.  Given that $1&lt;x^2-xy+y^2&lt;2$, how can I show that $\frac 29&lt;x^4+y^4&lt;8$ ?  Then can I use that to prove that for any natural number $n&gt;3$  $$x^{2^n}+y^{2^n}&gt;\frac 2{3^{2^n}} \text{?}$$
117251
I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!
102647
Prove the following, without induction. Is it possible?         The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.   And if $n$ is odd, it has only one real root.       I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$". But failed. What is the way to search the related topics?
102749
$$(z+1)^6=z^6 $$ $$\frac{(z+1)^6}{z^6 }=1$$ $$\left(\frac{(z+1)}{z }\right)^6=1$$ $$\left(\frac{(z+1)}{z }\right)^6=\cos(2*\pi*k)+i\sin(2*\pi*k)$$ $$\frac{(z+1)}{z }=\left(\cos(2*\pi*k)+i\sin(2*\pi*k)\right)^{1\over6}$$ $$1+\frac{1}{z }=\cos({2*\pi*k\over6})+i\sin({2*\pi*k\over6})$$  $$\frac{1}{z }=(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3})$$ $$\frac{1}{(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3}) }=z$$ I dont know how else to solve this. A lot of help is appreciated. Also i know that k should go from 1-5.
48493
I would appreciate any help with this problem: $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $  I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha to get the solutions so I know real ones exist (2 of them, actually), but I have no idea how to get them.
104814
Can all the roots of the polynomial equation (with real co-efficients) $a_nx^n+...+a_3x^3+x^2+x+1=0$ be real ? I tried using Vieta's formulae $\prod \alpha=\dfrac {(-1)^n}{a_n}$ , $(\prod \alpha )(\sum \dfrac 1 \alpha)=(-1)^{n-1}\dfrac 1 a_n$ , so $\sum \dfrac1\alpha = -1$ but this is not going anywhere , please help .
121323
If we have:  $a&gt;0,a+b+c \ge 0,a-b+c \ge 0,a-c \ge 0,b^2-4ac \ge 0$  Then Prove that there exists two roots in $[-1,1]$.  We have $f(1),f(-1) \ge 0$ so if we could prove there is a number $c$ that $f(c)&lt;0$ and $-1&lt;c&lt;1$ we are done but I can't do that.  Edit:$f(x)=ax^2+bx+c$  Edit 2 I made a typo and corrected it now.Now using answer of @George Law It only remains to check the case $\Delta =0$ which is likely to cause problem with $a-c \ge o$ because we didn't use it.
115629
Any help with the following question would be very much appreciated.     The equation $2x^2 + ax + (b + 3) = 0$ has real roots. Find the minimum value of $a^2 + b^2$.   Please note that, unfortunately, the question provides no clarification regarding the nature of $a$ and/or $b$.