Đáp án:
$6$ nghiệm
Giải thích các bước giải:
$\begin{array}{l} \sin^2x - \cos^23x = 0\\ \Leftrightarrow \dfrac{1 - \cos2x}{2} - \dfrac{1+ \cos6x}{2} = 0\\ \Leftrightarrow \cos2x + \cos6x =0\\ \Leftrightarrow 2\cos4x\cos2x = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos4x = 0\\\cos2x = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{4}\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2} \end{array}\right.\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ 0 < x < \pi\\ \Rightarrow \left[\begin{array}{l}0 < \dfrac{\pi}{8} + k\dfrac{\pi}{4} < \pi\\0 < \dfrac{\pi}{4} + k\dfrac{\pi}{2} < \pi\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}-\dfrac12 < k < \dfrac72\\-\dfrac12 < k < \dfrac32\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}k = \left\{0;1;2;3\right\}\\k = \left\{0;1\right\}\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=\dfrac{\pi}{8}\\x=\dfrac{3\pi}{8}\\x=\dfrac{5\pi}{8}\\x=\dfrac{7\pi}{8}\\x=\dfrac{\pi}{4}\\x=\dfrac{3\pi}{4}\end{array}\right.\\ \text{Vậy có 6 nghiệm thỏa yêu cầu bài toán} \end{array}$
