How to find root and solution of the questions as below?

[karimkhan]

in 13 how to deal with the problem,

and in 14 in both the case we get x=2 and 4, so can we say equalances is 2...

 

[pka]

in 13 how to deal with the problem..

In the first question you can see that \(\displaystyle x=0\) is a solution.
Are there others? If I were I would graph \(\displaystyle 2\cos\left(\frac{x^2+x}{6}\right)-2^x-2^{-x}\) to get some ideas.

 

[Deleted member 4993]

I do not know how somebody would answer question (1) without graphing.

Wolfram-alfa shows only one real root (x=0)