Finding all intersections of $f(x)= \sin(x)+1$ and $g(x)= \cos(x)$ on the interval $[0,4\pi]$ 
The question asks to find all the points where $f(x)= \sin(x)+1$ intersects with $g(x)= \cos(x)$ on the interval $[0,4\pi]$.

I started by setting both equations equal to each other resulting in the new equation:
$$\sin(x)+1 = \cos(x)$$
I thought that if I was somehow able to use trigonometric identities in order to make $\sin(x)$ and $\cos(x)$ end up multiplying to each other so that I don't get rid of any solutions and can solve more easily.
My process:
sin(x)+1 = cos(x)
(sin(x)-cos(x))^2= (-1)^2
sin^2(x)-2sin(x)cos(x)+cos^2(x)=1
sin^2(x)+cos^2(x)=1+2sin(x)cos(x)                 Pythagorean Identity
1= 1+2sin(x)cos(x)                                Subtract 1 from both sides
0= 2sin(x)cos(x)
This states that the solution is anytime cos(x) or sin(x) equals zero. This would mean x= 0,π/2,π,3π/2,2π,5π/2,3π,7π/2, and 4π.
But when I graphed this I got that the solutions are at: x=0,3π/2,2π,7π/2, and 4π. This is half of what I thought would be the solutions. 
I used logic to try to solve it now. 
I started again with setting the equations to each other again and then guessing and checking.
                                        sin(x)+1 = cos(x)  

I knew that for this to be true sin(x) would have to equal zero when cos(x) would have to equal one or sin(x) would have to equal negative one when cos(x) would have to equal zero.
This in mind. I listed all the places:
sin(x) equals zero: 0,π, and 2π
cos(x) equals one: 0, 2π
Where they coincided I knew there was a solution. Here two solution were 0 and 2π.
Then I did the same for when sin(x) equals negative one and cos(x) equals zero
sin(x) equals negative one: 3π/2
cos(x) equals zero: π/2, 3π/2
Here another solution was 3π/2.
Because sin and cos graphs oscillate I know that is I add 2π to every one of these solutions I would get the rest of the solutions from [2π,4π].
Although, when problems become more complicated I can't always rely on guess and check so I was wondering how I could algebraically solve it since I can't figure it out. 
 A: When you square both sides you run the risk of introducing false solutions.
$(\cos x - \sin x) = 1$
Squaring both sides...$(\cos x - \sin x)^2 = 1$ will now gives a "solution" when $(\cos x - \sin x) = -1$
So, if this is the route you take, you must be careful to check which of your solutions are associated with which equation.
When $\sin x > 0$ then $\sin x + 1 > 1$ and it is always the case that $\cos x \le 1.$  Similarly when $\cos x<0$ there is impossible for $\sin x + 1$ to be less than $0.$  We can use these fact to eliminate the "extra" solutions.
An alternative approach is to say
$\sqrt 2 (\frac {\sqrt 2}{2}\cos x - \frac {\sqrt 2}{2}\sin x) =1\\
\sqrt 2 (\cos \frac {\pi}{4}\cos x - \sin\frac {\pi}{4}\sin x) =1\\
\cos (x+\frac {\pi}{4}) = \frac {\sqrt 2}{2}$
A: Factorize the equation $f(x)=g(x)$ instead
\begin{align}
\sin x+1 - \cos x & = 2\sin\frac x2\cos\frac x2+2\sin^2\frac x2\\
& = 2\sin\frac x2(\cos\frac x2+ \sin\frac x2)\\
&= 2\sqrt2 \sin\frac x2\cos(\frac x2-\frac\pi4)=0
\end{align}
which leads to $\sin\frac x2=0$ and $\cos(\frac x2+\frac\pi4)=0$. Thus, the intersections over $[0,4\pi]$ are $x= 0, 2\pi, 4\pi, \frac{3\pi}2, \frac{7\pi}2$.
