Quay lại đáp án

b)Phương trình tương đương với:$\sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}\sin\left (\frac{9\pi}{10}-\frac{3x}{2} \right )$Đặt: $\frac{3\pi}{10}-\frac{x}{2}=t$, ta có: $\sin t=\frac{1}{2}\sin3t$$\Leftrightarrow 2\sin t=3\sin t-4\sin^3t$$\Leftrightarrow \left[\begin{array}{l} \sin t=0\\\sin t=\frac{1}{2}\\\sin t=\frac{-1}{2} \end{array} \right.$$\Leftrightarrow \left[ \begin{array}{l} t=k\pi\\t=\pm\frac{\pi}{6}+k2\pi\\t=\pm\frac{5\pi}{6}+k2\pi \end{array} \right.,k\in\mathbb{Z}$$\Leftrightarrow \left[ \begin{array}{l} x=\frac{3\pi}{5}-k\pi\\x=\frac{4\pi}{15}-k4\pi\\x=\frac{14\pi}{15}-k4\pi \end{array} \right.,k\in\mathbb{Z}$

c)Phương trình tương đương với:$\sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}\sin\left (\frac{9\pi}{10}-\frac{3x}{2} \right )$Đặt: $\frac{3\pi}{10}-\frac{x}{2}=t$, ta có: $\sin t=\frac{1}{2}\sin3t$$\Leftrightarrow 2\sin t=3\sin t-4\sin^3t$$\Leftrightarrow \left[\begin{array}{l} \sin t=0\\\sin t=\frac{1}{2}\\\sin t=\frac{-1}{2} \end{array} \right.$$\Leftrightarrow \left[ \begin{array}{l} t=k\pi\\t=\pm\frac{\pi}{6}+k2\pi\\t=\pm\frac{5\pi}{6}+k2\pi \end{array} \right.,k\in\mathbb{Z}$$\Leftrightarrow \left[ \begin{array}{l} x=\frac{3\pi}{5}-k\pi\\x=\frac{4\pi}{15}-k4\pi\\x=\frac{14\pi}{15}-k4\pi \end{array} \right.,k\in\mathbb{Z}$