Gỉai pt 6cos^2x + 5sinx -7 =0

Đáp án:

$\left\{ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right).$

Lời giải:

\(\begin{array}{l} 6{\cos ^2}x + 5\sin x - 7 = 0\\ \Leftrightarrow 6\left( {1 - {{\sin }^2}x} \right) + 5\sin x - 7 = 0\\ \Leftrightarrow 6 - 6{\sin ^2}x + 5\sin x - 7 = 0\\ \Leftrightarrow 6{\sin ^2}x - 5\sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \dfrac{1}{2}\\ \sin x = \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right). \end{array}\)

Vậy $\left\{ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right).$