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Nghiệm của pt: 2sin^2 x – 3sin x + 1 = 0 thỏa mãn 0 <= x <= pi/2 là?
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$2\sin^2x-3\sin x+1=0$
$\Leftrightarrow \sin x=1$, $\sin x=\dfrac{1}{2}$
$+) \sin x=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi$
$x\in[0;\dfrac{\pi}{2}]\Rightarrow x=\dfrac{\pi}{2}$
$+) \sin x=0,5\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi$ hoặc $x=\dfrac{5\pi}{6}+k2\pi$
$x\in[0;\dfrac{\pi}{2}\Rightarrow x=\dfrac{\pi}{6}$
Vậy $S=\{\dfrac{\pi}{2};\dfrac{\pi}{6}\}$
Đáp án:
Giải thích các bước giải: bấm máy mode 5 3 nhé