giải phương trình dựa vào góc liên kết
a)sin3x=-sinx
b)cos2x=-cosx
c)cos3x=sin2x
d)cos7x=-sin3x
e)sin(2x+pi/6)=-cos(x+pi/3)
f)cos(3x-pi/4)=sin(x-pi/3)
giải phương trình dựa vào góc liên kết
a)sin3x=-sinx
b)cos2x=-cosx
c)cos3x=sin2x
d)cos7x=-sin3x
e)sin(2x+pi/6)=-cos(x+pi/3)
f)cos(3x-pi/4)=sin(x-pi/3)
$\begin{array}{l}a)\,\sin3x=-\sin x\\ \Leftrightarrow \sin3x=\sin(-x)\\ \Leftrightarrow \left[ \begin{array}{l}3x = -x + k2\pi\\3x = \pi + x + k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x=k\dfrac{\pi}{2}\\3x = \dfrac{\pi}{2} + k\pi\end{array} \right.\quad (k \in \Bbb Z)\\ b)\,\cos2x=-\cos x\\ \Leftrightarrow \cos2x=\cos(\pi -x)\\ \Leftrightarrow \left[ \begin{array}{l}2x=\pi – x + k2\pi\\2x = x – \pi + k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{3} + k\dfrac{2\pi}{3}\\x = – \pi + k2\pi\end{array} \right.\quad (k \in \Bbb Z)\\ c)\,\cos3x=\sin2x\\ \Leftrightarrow \cos3x=\cos\left(\dfrac{\pi}{2} -2x\right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \dfrac{\pi}{2} -2x + k2\pi\\3x = 2x – \dfrac{\pi}{2}+ k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{10} + k\dfrac{2\pi}{5}\\x= – \dfrac{\pi}{2}+ k2\pi\end{array} \right.\quad (k \in \Bbb Z)\\ d)\,\cos7x=-\sin3x\\ \Leftrightarrow \cos7x=\sin(-3x)\\ \Leftrightarrow \cos7x=\cos\left(\dfrac{\pi}{2}+3x\right)\\ \Leftrightarrow \left[ \begin{array}{l}7x = \dfrac{\pi}{2}+3x + k2\pi\\7x = -\dfrac{\pi}{2}-3x+ k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{20}+ k\dfrac{\pi}{5}\end{array} \right.\quad (k \in \Bbb Z)\\ e)\,\sin\left(2x+\dfrac{\pi}{6}\right)=\cos\left(\pi +x+\dfrac{\pi}{3}\right)\\ \Leftrightarrow \sin\left(2x+\dfrac{\pi}{6}\right)=\sin\left(\dfrac{\pi}{2} – \pi -x-\dfrac{\pi}{3}\right)\\ \Leftrightarrow \sin\left(2x+\dfrac{\pi}{6}\right)=\sin\left(-\dfrac{5\pi}{6} -x\right)\\ \Leftrightarrow \left[ \begin{array}{l}2x+\dfrac{\pi}{6} = -\dfrac{5\pi}{6} -x + k2\pi\\2x+\dfrac{\pi}{6} = \dfrac{11\pi}{6} +x+ k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = -\dfrac{\pi}{3} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{3} + k2\pi\end{array} \right.\quad (k \in \Bbb Z)\\ f)\,\cos\left(3x-\dfrac{\pi}{4}\right)=\sin\left(x-\dfrac{\pi}{3}\right)\\ \Leftrightarrow \cos\left(3x-\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{2} -x+\dfrac{\pi}{3}\right)\\ \Leftrightarrow \cos\left(3x-\dfrac{\pi}{4}\right)=\cos\left(\dfrac{5\pi}{6}-x\right)\\ \Leftrightarrow \left[ \begin{array}{l}3x-\dfrac{\pi}{4} = \dfrac{5\pi}{6}-x + k2\pi\\3x-\dfrac{\pi}{4} = x-\dfrac{5\pi}{6}+ k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x= \dfrac{13\pi}{48} + k\dfrac{\pi}{2}\\x= -\dfrac{7\pi}{24}+ k\pi\end{array} \right.\quad (k \in \Bbb Z)\\ \end{array}$