Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$ 
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$

Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}$
Why is this solution wrong? Is there more than one $x$ value for $\sin x - 1=0$ or $(\sqrt{2} \cos x +1)=0$ in the given interval? How do I know that?

Your overlooked the fact that there are two values of $x \in [0, 2\pi]$ such that $\sqrt{2}\cos x + 1 = 0$.
As we can see from the sine graph, there is only one value of $x \in [0, 2\pi]$ such that $\sin x = 1$.

A number $x$ is a solution of the equation $\sqrt{2}\cos x + 1 = 0 \iff \cos x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$.  As we can see from the cosine graph, since
$$-1 < -\frac{\sqrt{2}}{2} < 1$$
there are two values of $x$ in the interval $[0, 2\pi]$ such that $\cos x = -\frac{\sqrt{2}}{2}$.

The interval $[0, 2\pi]$ contains the coterminal angles $0$ and $2\pi$, so there are two values of $x$ such that $\cos x = 1$, namely $0$ and $2\pi$.
However, if we were to instead consider the interval $[0, 2\pi)$, the equations $\sin x = 1$ and $\sin x = -1$ each have one solution, while the equation $\sin x = y$ has two solutions whenever $-1 < y < 1$.  Similarly, in the same interval, the equations $\cos x = 1$ and $\cos x = -1$ each have one solution, while the equation $\cos x = y$ has two solutions whenever $-1 < y < 1$.
Next, consider the diagram below.

When does $\sin\theta = \sin\varphi$?
Two angles have the same sine if the $y$-coordinates of the points where their terminal sides intersect the unit circle are equal.  Clearly, this is true if $\theta = \varphi$.  By symmetry, it is also true when $\theta = \pi - \varphi$.  Moreover, it is also true for any angles coterminal with these angles.  Hence, the general solution of the equation $\sin\theta = \sin\varphi$ is
\begin{align*}
\theta & = \varphi + 2k\pi, k \in \mathbb{Z} & \theta = \pi - \varphi + 2m\pi, m \in \mathbb{Z}
\end{align*}
Now, consider the equation $\sin x - 1 = 0$.
\begin{align*}
\sin x - 1 & = 0\\
\sin x & = 1
\end{align*}
A particular solution of this equation is
$$x = \arcsin(1) = \frac{\pi}{2}$$
Thus, the general solution of this equation is
\begin{align*}
x & = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & x & = \pi - \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}\\
  & & & = \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}
\end{align*}
Since we want a solution in the interval $[0, 2\pi]$, we must take $k = 0$ and $m = 0$, which yields the unique solution
$$x = \frac{\pi}{2}$$
When is $\cos\theta = \cos\varphi$?
Two angles have the same cosine if the $x$-coordinates of the points where their terminal sides intersect the unit circle are equal.  Clearly, this is true if $\theta = \varphi$.  By symmetry, it is also true if $\theta = -\varphi$.  Moreover, any angles coterminal with these angles have the same cosine.  Thus, the general solution of the equation $\cos\theta = \cos\varphi$ is
\begin{align*}
\theta & = \varphi + 2k\pi, k \in \mathbb{Z} & \theta = -\varphi + 2m\pi, m \in \mathbb{Z}
\end{align*}
Now, consider the equation $\sqrt{2}\cos x + 1 = 0$.
\begin{align*}
\sqrt{2}\cos x + 1 & = 0\\
\sqrt{2}\cos x & = -1\\
\cos x & = -\frac{1}{\sqrt{2}}\\
\cos x & = -\frac{\sqrt{2}}{2}
\end{align*}
A particular solution of this equation is
$$x = \arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$$
Thus, the general solution of the equation $\sqrt{2}\cos x + 1 = 0$ is
\begin{align*}
x & = \frac{3\pi}{4} + 2k\pi, k \in \mathbb{Z} & x & = -\frac{3\pi}{4} + 2m\pi, m \in \mathbb{Z}
\end{align*}
Since we seek solutions in the interval $[0, 2\pi]$, we must take $k = 0$ and $m = 1$, which yields the solutions
\begin{align*}
x & = \frac{3\pi}{4} & x & = 2\pi - \frac{3\pi}{4} = \frac{5\pi}{4}
\end{align*}
