Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{\frac{x}{2}}\cos{\frac{x}{2}} + asin^2{\frac{x}{2}} + sin^2{\frac{x}{2}} + a\cos^2{\frac{x}{2}} - \cos^2{\frac{x}{2}} = 1$$
$$ a\left(sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 = 1 - \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}$$
$$ a\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 =2\cos^2{\frac{x}{2}}$$
I can't finish...
 A: $$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
$$ a\sin{x}+a\left(\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}\right)-\left(\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}\right)=1 $$ $$ a\sin x+a-\cos x=1 $$
$$ a\sin x-\cos x=(1-a) $$ $$ \frac{a}{\sqrt{1+a^2}}\sin x-\frac{1}{\sqrt{1+a^2}}\cos x=\frac{(1-a)}{\sqrt{1+a^2}} $$ $$\sin x\cos \alpha-\cos x\sin \alpha=\sin \beta $$ Where $\alpha=\cos^{-1}\left(\frac{a}{\sqrt{1+a^2}}\right)$ & $\beta=\sin^{-1}\left(\frac{1-a}{\sqrt{1+a^2}}\right)$ $$\sin(x-\alpha)=\sin \beta$$ $$x-\alpha=2n\pi+\beta\iff x=2n\pi+\alpha+\beta$$ or $$x-\alpha=(2n+1)\pi-\beta\iff x=(2n+1)\pi+\alpha-\beta$$
A: HINT....The equation simplifies to $ a\sin x +a- \cos x=1$ can you take it from there?
A: $$ a\sin x + a(\sin^2 (x/2) + \cos^2 (x/2)) + \sin^2 (x/2) - \cos^2 (x/2) =1$$
$$ a\sin x + a(\sin^2 (x/2) + \cos^2 (x/2)) + 1 - 2\cos^2 (x/2) =1$$
$$ a \sin x + a - \cos x =1$$
A: Note that your equation reduces to
\begin{align}
1
&= a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} \\
&= a \sin x +a \left(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}\right) +
\sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a + \sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a - \cos(x) \\
\end{align}
And this could be rewritten as
$$
\frac{1 + \cos x}{1 + \sin x} = a 
$$
for $\sin x \ne -1$ or 
$$
\cos x = -1 
$$
for $\sin x = -1$ which can not happen.
So we assume $\sin x \ne -1$.
