Proving a trigonometric inequality $(1-\sin a)x^2 -2x\cos a + 1+ \sin a \ge 0$ $(1-\sin (a))x^2 -2x\cos(a) + 1+ \sin( a) \ge 0$, where $a,x$ are any two real constants.
Any suggestions on how to prove this ? I tried playing with it, but nothing useful came out.
 A: Yet another way, by AM-GM
$$(1-\sin a) x^2 + (1+\sin a) \ge 2\sqrt[]{(1-\sin^2 a)x^2}= 2\lvert x \cos a \rvert$$
A: Let $$x^2(1-\sin a)-2x\cos a+(1+\sin a)=y$$
$$\iff x^2(1-\sin a)-2x\cos a+(1+\sin a)-y=0$$
For real $x,$
$$(2\cos a)^2-4(1-\sin a)(1+\sin a-y)\ge0$$
As $\displaystyle\cos^2a=(1-\sin a)(1+\sin a)$ and $\displaystyle4(1-\sin a)\ge0, $
$\displaystyle(1+\sin a)-(1+\sin a-y)\ge0\iff y\ge0$
A: If $1-\sin a=0$, then $\sin a=1$ and $\cos a=0$ so that the problem is trivial. If $1-\sin a\neq 0$, then the given expression is a polynomial of the form $dx^2+bx+c$ where $b^2=4dc$. So the polynomial factors into $d\left(x+\frac{b}{2d}\right)^2\geq 0$ (note that in this case $d=1-\sin a\geq 0$).
A: Set $\sin a=\frac{2t}{1+t^2},\cos a=\frac{1-t^2}{1+t^2}, t=\tan(a/2)$, then the original inequality becomes:
$$\frac{(1 + t - x +  tx)^2}{ (1 + t^2)}\ge 0$$
Which obviously holds.
A: $\cos^2 a = 1-\sin^2a=(1+\sin a)(1-\sin a)$
$1+\sin a = \frac {\cos^2 a}{1-\sin a}$
Substituting this value in the equation -- 
$(1-\sin a)x^2 -2x\cos(a) + [1+ \sin( a)]  $ 
=$(1-\sin a)x^2 -2x\cos(a) + \frac {\cos^2 a}{1-\sin a}  $
=$[x \root  \of {1-\sin a} - \frac {\cos a}{\root \of {1-\sin a}}  ]^2$
$\ge 0$
A: This is a degree 2 polynomial in $x$ with discriminant proportional to $(\cos a)^2 - (1-\sin a)(1+\sin a) = 0$, hence it keeps a constant sign.
