[TEX]t =sinx+\sqrt{2-sin^2x} DK: -2 \le t \le 2 \\ \Rightarrow t^2 - 2 = 2 sinx\sqrt{2-sin^2x}[/TEX]
Thế vào ta có :
[TEX] 2t + t^2 - 2 = 6[/TEX]
[TEX]\Leftrightarrow t^2 + 2t - 8 = 0 [/TEX]
[TEX]\Leftrightarrow \left[ t = - 4 (loai) \\ t = 2 (thoa) [/TEX]
[TEX]\Leftrightarrow sin x + \sqrt{2-sin^2x} = 2[/TEX]
[TEX]VT \le VP \\ "=" \Leftrightarrow sin x = \sqrt{2-sin^2 x } [/TEX]
[TEX]\Leftrightarrow sin x = 1 [/TEX]