Đáp án:
$9$ nghiệm
Giải thích các bước giải:
$2\cos^2x +\sin x = 2$
$\Leftrightarrow 2(1-\sin^2x)+\sin x - 2 = 0$
$\Leftrightarrow 2\sin^2x -\sin x = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x =\dfrac12\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x =\dfrac{\pi}{6} + k'2\pi\\x =\dfrac{5\pi}{6} + k''2\pi\end{array}\right.\quad (k, k', k''\in\Bbb Z)$
Ta có:
$\quad 0 \leq x \leq 4\pi$
$\Leftrightarrow \left[\begin{array}{l}0≤ k\pi≤4\pi\\0\leq\dfrac{\pi}{6} + k'2\pi\leq 4\pi\\0\leq\dfrac{5\pi}{6} + k''2\pi\leq 4\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}0\leq k \leq 4\\-\dfrac{1}{12}\leq k'\leq \dfrac{23}{12}\\-\dfrac{5}{12}\leq k''\leq \dfrac{19}{12}\end{array}\right.$
$\Rightarrow \left[\begin{array}{l}k =\left\{0;1;2;3;4\right\}\\k'=\left\{0;1\right\}\\k''=\left\{0;1\right\}\end{array}\right.$
$\Rightarrow 9$ giá trị $k, k', k''$ tương ứng $9$ nghiệm thoả mãn đề bài
