ĐK: $-1\leq x\leq 1$Đặt $x=sint,t\in [-\frac{\pi}{2};\frac{\pi}{2}]\Rightarrow \sqrt{1-x^2}=\sqrt{1-sin^2t}=\left| {cost} \right|=cost$
Pt $\Leftrightarrow \sqrt{2(1-cost)}=sint(1+cost)$
$\Leftrightarrow \sqrt{4sin^2\frac{t}{2}}=2sint.cos^2\frac{t}{2}$
$\Leftrightarrow \left| {sin\frac{t}{2}} \right|=2sin\frac{t}{2}.cos^3\frac{t}{2} (1)$
- Với $t\in [-\frac{\pi}{2};0)\Rightarrow \frac{t}{2}\in [-\frac{\pi}{4};0)\Rightarrow \begin{cases}sin\frac{t}{2}<0 \\ cos\frac{t}{2}>0 \end{cases}$ thì $(1)\Leftrightarrow -sin\frac{t}{2}=2sin\frac{t}{2}cos^3\frac{t}{2}\Leftrightarrow sin\frac{t}{2}(1+2cos^3\frac{t}{2})=0 (VN)$
- Với $t\in [0;\frac{\pi}{2}]\Rightarrow \frac{t}{2}\in [0;\frac{\pi}{4}]\Rightarrow \begin{cases}sin\frac{t}{2}\geq 0\\ cos\frac{t}{2}\leq 0\end{cases}$ thì $(1)\Leftrightarrow sin\frac{t}{2}=2sin\frac{t}{2}cos^3\frac{t}{2}\Leftrightarrow sin\frac{t}{2}(1-2cos^3\frac{t}{2})=0\Leftrightarrow sin\frac{t}{2}=0\vee cos\frac{t}{2}=\frac{1}{\sqrt[3]{2}}$
- Với $\begin{cases}sin\frac{t}{2}=0 \\ t\in [0;\frac{\pi}{2}]\end{cases}\Rightarrow \begin{cases}t=k2\pi \\ t\in[0;\frac{\pi}{2}] \end{cases}\Leftrightarrow t=0\Leftrightarrow x=sint=0$
- Với $\begin{cases}cos\frac{t}{2}=\frac{1}{\sqrt[3]{2}} \\ t\in [0;\frac{\pi}{2}]\end{cases}\Rightarrow \begin{cases}sin\frac{t}{2}=\sqrt{1-\frac{1}{\sqrt[3]{4}}}=\frac{\sqrt{\sqrt[3]{4}-1}}{\sqrt[3]{2}} \\ t\in [0;\frac{\pi}{2}]\end{cases}\Leftrightarrow x=sint=2sintcost=\sqrt{\sqrt[3]{4}-1}$
Vậy $x=0\vee x=\sqrt{\sqrt[3]{4}-1}$ là nghiệm của pt