Trigonometric equation $\sin x+1=\cos x$ $$\sin x+1=\cos x,\quad x\in[-\pi,\pi]$$
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?
 A: This is an instance of a “linear equation in sine and cosine”. There are several methods for solving them.
First method.
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 $x=-\pi/2$.
Second method
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 $x=-\pi/2$.
Third method
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
$$
-\pi\le x\le \pi
$$
A: 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?
A: HINT:
$$1+\sin(x)=\cos(x)\Longleftrightarrow$$
$$1-\cos(x)+\sin(x)=0\Longleftrightarrow$$
$$1-\sqrt{2}\left(\frac{\cos(x)}{\sqrt{2}}-\frac{\sin(x)}{\sqrt{2}}\right)=0\Longleftrightarrow$$
$$1-\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)\cos(x)-\sin\left(\frac{\pi}{4}\right)\sin(x)\right)=0\Longleftrightarrow$$
$$1-\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=0\Longleftrightarrow$$
$$-\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=-1\Longleftrightarrow$$
$$\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)=1\Longleftrightarrow$$
$$\cos\left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}}$$
A: 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 $0 + 1 \neq -1$
A: rewrite your equation into the form
$$2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
 \right)  \right) ^{2}}}+1-{\frac {1- \left( \tan \left( x/2 \right) 
 \right) ^{2}}{1+ \left( \tan \left( x/2 \right)  \right) ^{2}}}
=0$$
A: $$\begin{align}
\sin(x)+1&=\cos(x)\\
\sin(x)-\cos(x)&=-1\\
(\sin(x)-\cos(x))^2&=(-1)^2\\
\sin^2(x)+\cos^2(x)-2\sin(x)\cos(x)&=1\\
1-2\sin(x)\cos(x)&=1\\
\sin(x)\cos(x)&=0\\
\end{align}$$
Then you can test $x=-\pi, -\pi/2, 0, \pi/2, \pi$.
A: $\sin x + 1 = \cos x$
$\sin \pm \pi + 1 = 1 \ne \cos \pm \pi = -1$ 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 $0$.
