Find number of real solutions of the equation $x+1=99\sin (\pi (x+1))$ I think this question is mostly about having a good eye on specific properties. Here is my logic 
Since the term on the RHS is $99\sin (\pi(x+1))$. For any integer values of $x$, the term will end up being 
$99\sin (n\pi)$, which is obviously zero. That gives us one solution. 
Now if we go for non integer solutions, the possibilities seem endless. I am might not know some properties required to sovle this question, so please help me with that. 
 A: I think the best strategy for questions like these are to look at the properties of their graphs. If we bring $99$ from the RHS to the LHS, we have:
$\frac{(x + 1)}{99} = sin(\pi(x + 1)) = sin(\pi x + \pi)$
Now it is evident from the LHS of the above equation, that we're trying to draw a line $y = \frac{x}{99} + \frac{1}{99}$ with $y$-intercept of $\frac{1}{99}$. Also, our RHS is bounded between $1$ and $-1$. So for $x \geq 99$ and $x < -100$, we will definitely have no solution. That's enough with the LHS for now, let's shift our focus to RHS. We have a sine wave which is positively shifted by $\pi$, and covers a period in $[x, x + 2]$. Moreover, in a period, a sine wave covers every value between $[-1, 1]$ so it will intersect the line exactly twice in a period. This is a major clue, we just need to now look carefully at how many times those periods would be completed in the positive and the negative $x$-axis given the above constraints, and we'll have our answer.
Thus for the positive $x$-axis we'll have about $\lfloor \frac{99}{2} \rfloor * 2 = 98$ real solutions. For the negative $x$-axis, similarly, $\lfloor \frac{-100}{2} \rfloor * 2 = 100$ for a total of $198 + 1$ real solutions counting the one additional solution due to $x = 0$.
A: By a shift in the equation, it will be altered to $$x=99\sin \pi x$$which has $x=0$ as one of its roots. Since this equation is symmetric w.r.t. the origin, the number of its roots will become:$$N=2\times \text{Number of positive roots} +1$$hence we only try to find the number of its positive roots.
Clearly $x$ and $99\sin \pi x$ coincide for $x>0$ only when $\sin \pi x>0$, or $$2k <x<2k+1$$ and $x\le 99$. This yields us a maximal value for $k$ for which we must have $$2k \le 99\\2k+1\le 99$$which gives us $k\le 49$. Since the first $k=49$ positive peaks of $99\sin \pi x$ coincide with $x$, each of which meeting it twice. For $x\ge 2\times 49+1=99$ however, there is no answer since we must have $$x>2\times 49+2 >99$$hence the total number of positive roots becomes $99$ and the total number of roots will then be $199$.
A: For generality, assume the equation is 
$$t=a\sin t,\>\>\>a>1$$
The number of roots is given analytically by 
$$N(a)= 3+4\cdot \text{Int}\left[\frac1{2\pi\csc^{-1}a}-\frac14\right]\tag1$$
where Int[] denotes integer values. So, for the equation under consideration $x+1=99\sin (\pi (x+1))$, we have $a=99\pi$. Plug into (1) to obtain the number of roots $N(99\pi)=199$.
PS: For the derivation of (1), refer to What is the number of solutions of $x = a\sin(x)$
A: A simplier approach
The number of real solution to the equation $x+1 = 99\sin(\pi(x+1))$
Say $x+1 = n$
The equation becomes $n = 99\sin(\pi*n)$
$n/99 = \sin(\pi*n)$
$\arcsin(n/99) = \pi*n$
$\arcsin(m) = 99\pi*m$
Where $n= 99m$
Remember that $\arcsin(m)$ is real if only $m ≤ 1$ and $m ≥ -1$
Therefore $-1 ≤ m ≤ 1$
$-1 ≤ n/99 ≤ 1$
$-1 ≤ (x+1)/99 ≤ 1$
$-100 ≤ x ≤ 98$
This is the range of values $x$ must take for a real solution to occur
The graph of our function $x+1 = 99\sin(\pi(x+1))$ is offcourse periodic because of the $\sin$, the straight line cuts it at every linear displacement
So the number of real roots here is the number of elements in the set of $x$, where $-100 ≤ x ≤ 98$
Counting all the numbers here including 0 gives us $199$
