$$\sin x+1=\cos x,\quad x\in[-\pi,\pi]$$
How do you solve by squaring both sides? the solution is $x\in\{-\pi/2,0\}$ so the solutions $\pi$ and $-\pi$ are inadmissible, I do not understand how by subbing $-\pi$ back into both sides of the equations makes them unequal, and the same for positive $\pi$. Which equation are you subbing $\pi$ into to check, the original?
This is an instance of a “linear equation in sine and cosine”. There are several methods for solving them.
Write $X=\cos x$, $Y=\sin x$ and consider the system $$ \begin{cases} 1+Y=X\\ X^2+Y^2=1 \end{cases} $$ Substitute in the second equation to get $$ 1+2Y+Y^2+Y^2=1 $$ which gives $$ Y^2+Y=0 $$ so $Y=0$ or $Y=-1$. Thus we get the two solutions $$ \begin{cases} X=1\\ Y=0 \end{cases} \qquad\text{or}\qquad \begin{cases} X=0\\ Y=-1 \end{cases} $$ Solved with respect to $x\in[-\pi,\pi]$, they give $x=0$ or $x=-\pi/2$.
Rewrite the equation as $$ \cos x-\sin x=1 $$ and try to rewrite this as $A(\cos x\cos\phi-\sin x\sin\phi)=1$, with $A>0$. Thus we need $$ A\cos\phi=1,\qquad A\sin\phi=1 $$ so $A^2\cos^2\phi+A^2\sin^2\phi=2$, or $A^2=2$; thus $A=\sqrt{2}$ and $$ \cos\phi=\frac{1}{\sqrt{2}},\quad\sin\phi=\frac{1}{\sqrt{2}} $$ that is, $\phi=\pi/4$. Thus the equation becomes $$ \sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=1 $$ that means $$ x+\frac{\pi}{4}=\frac{\pi}{4}+2k\pi \qquad\text{or} x+\frac{\pi}{4}=-\frac{\pi}{4}+2k\pi $$ and we get again $x=0$ or $x=-\pi/2$.
Set $t=\tan(x/2)$ and recall that $$ \cos x=\frac{1-t^2}{1+t^2},\qquad \sin x=\frac{2t}{1+t^2} $$ that transforms the equation into $$ 1+\frac{2t}{1+t^2}=\frac{1-t^2}{1+t^2} $$ that becomes $$ 1+t^2+2t=1-t^2 $$ or $$ t^2+t=0 $$ so $t=0$ or $t=-1$. The first solution corresponds to $$ \frac{x}{2}=k\pi \to x=2k\pi $$ and the second solution corresponds to $$ \frac{x}{2}=-\frac{\pi}{4}+k\pi \to x=-\frac{\pi}{2}+2k\pi $$ and, again, in the given interval we have $x=0$ or $x=-\pi/2$.
You seem to have misunderstood what $x\in[-\pi,\pi]$ means. It means that you have to find all solutions $x$ such that $$ -\pi\le x\le \pi $$
Square both sides to get:
$$\sin^2 x + 2 \sin x + 1 = \cos^2 x.$$
Then use the identity $\sin^2 x + \cos^2 x = 1$ to eliminate the $\cos^2 x$ and you have a quadratic equation in $\sin x$.
Can you take it from here?
HINT:
$$1+\sin(x)=\cos(x)\Longleftrightarrow$$ $$1-\cos(x)+\sin(x)=0\Longleftrightarrow$$ $$1-\sqrt{2}\left(\frac{\cos(x)}{\sqrt{2}}-\frac{\sin(x)}{\sqrt{2}}\right)=0\Longleftrightarrow$$ $$1-\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)\cos(x)-\sin\left(\frac{\pi}{4}\right)\sin(x)\right)=0\Longleftrightarrow$$ $$1-\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=0\Longleftrightarrow$$ $$-\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=-1\Longleftrightarrow$$ $$\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=1\Longleftrightarrow$$ $$\cos\left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}}$$
To check the given solution, you check that $\sin (-\frac \pi 2) +1 = \cos (-\frac \pi 2)$, which is correct-$-1+1=0$. If you check $\pi$, you are asking whether $\sin \pi + 1 = \cos \pi$, which is not correct because $0 + 1 \neq -1$
rewrite your equation into the form $$2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}+1-{\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} =0$$
$$\begin{align} \sin(x)+1&=\cos(x)\\ \sin(x)-\cos(x)&=-1\\ (\sin(x)-\cos(x))^2&=(-1)^2\\ \sin^2(x)+\cos^2(x)-2\sin(x)\cos(x)&=1\\ 1-2\sin(x)\cos(x)&=1\\ \sin(x)\cos(x)&=0\\ \end{align}$$ Then you can test $x=-\pi, -\pi/2, 0, \pi/2, \pi$.
$\sin x + 1 = \cos x$
$\sin \pm \pi + 1 = 1 \ne \cos \pm \pi = -1$ so $\pm \pi$ are not solutions.
$\sin -\pi/2 + 1 = -1 + 1 = 0 = \cos -\pi/2 $ so $-\pi/2$ is solution.
$\sin 0 + 1 = 0 + 1 = \cos 0 $ so $0$ is solution.
But how to solve?
$\sin x + 1 = \cos x$
$\sin x - \cos x = -1$
$(\sin x -\cos x)^2 = (-1)^2$
$\sin^2 x + \cos^2 - 2\cos x \sin x = 1$
$1 - 2\cos x \sin x = 1$
$\cos x \sin x = 0$
So $\cos x = 0 $ or $\sin x = 0$.
So $x =\pm \pi/2$ or $x = \{0, \pi\}$.
But note.
$\sin x + 1 = \cos x \le 1$ so $\sin x = \cos x - 1 \le 0$ and $\cos x = \sin x + 1 \ge -1 + 1 = 0$.
$x = \pi/2 \implies \sin x = 1 > 0$ so $-\pi/2$ is not solution.
$x = \pi \implies \cos x = -1 < 0$ so $\pi$ is not solution
Solutions are $-\pi/2$ or $0$.
