Trigonometric inequality $\cos x+ \sin x>0$ Solve the inequality: $\cos x+ \sin x >0$ 
Why can't I square this to get $\sin 2x>0$? And what is the first step here then?
 A: Hint. Consider that 
$$\sin(x+\pi/4)=\sin(x)\cos(\pi/4)+\cos(x)\sin(\pi/4).$$
A: If you simply square the inequality, you lose information because
$$a>0\implies a^2>0$$ but the implied inequality is also true for $a<0$ !
Anyway, this can help you to find the roots by
$$(\cos x+\sin x)^2=1+\sin 2x=0\implies 2x=\frac{3\pi}2+2k\pi$$ or
$$x=\frac{3\pi}4+k\pi.$$
Then as these roots are simple for the original function, the sign alternates between the roots and it is positive in the ranges
$$(-\frac\pi4+2k\pi,\frac{3\pi}4+2k\pi)$$

A: Hint: Try to express  your function as $A\sin(x+x_0)$.
$$\cos(x)+\sin(x)=A\sin(x+x_0)=A\sin(x)\cos(x_0)+A\cos(x)\sin(x_0)$$
By comparing both expressions you will get 
$$1=A\sin(x_0)$$
$$1=A\cos(X_0)$$
Dividing both expressions gives you $1=\tan(x_0) \implies x_0 = \pi/4$ or $x_0 = -\pi/4$. And squaring and adding both gives you $1=A^2 \implies A =\pm 1$
Check which combination of $x_0$ and $A$ leads to a correct result.
A: Squaring an inequation $f(x)>0$ is certainly a bad idea to solve it, because $f(x)^2>0$ is true for every $x$ that is not a zero of the function $f$. $A^2>0$ does not imply, at all, that $A>0$.
To solve the inequation $\sin x+\cos x>0$, I'd use some elementary calculus. Define $f(x)=\sin x+\cos x$. This function is continuous, so finding its zeros seems a good idea. They are $3\pi/4+n\pi$, $n\in\Bbb Z$. For $x=2n\pi$, $n\in\Bbb Z$, $f(x)>0$, and for $x=(2n+1)\pi$, $n\in\Bbb Z$ we have $f(x)<0$.
The solution is the union of the intervals
$$(-\pi/4+2n\pi,3\pi/4+2n\pi)$$
for $n\in\Bbb Z$.
A: $$\cos x+ \sin x >0\Rightarrow \frac{\cos}{\sqrt2}+\frac{\sin x}{\sqrt2}=\sin(\frac{\pi}{4}+x)\gt0$$
It follows $$x\in\bigcup\space\left]-\frac{\pi}{4}+2k\pi,\frac{3\pi}{4}+2k\pi\space\right[$$
