b.
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=-\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=-\pi+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\dfrac{3}{5}sinx-\dfrac{4}{5}cosx=1\)
Đặt \(\dfrac{3}{5}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{4}{5}=sina\)
Pt trở thành:
\(sinx.cosa-cosx.sina=1\)
\(\Leftrightarrow sin\left(x-a\right)=1\)
\(\Leftrightarrow x-a=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=a+\dfrac{\pi}{2}+k2\pi\)
d.
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx-\dfrac{\sqrt{2}}{2}cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
e.
Nhân 2 vế với \(\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}cosx+\dfrac{\sqrt{2}}{2}sinx=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=cos\left(\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{\pi}{12}+k2\pi\end{matrix}\right.\)
f.
\(\Leftrightarrow\dfrac{12}{13}cos2x+\dfrac{5}{13}sin2x=1\)
Đặt \(\dfrac{12}{13}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{5}{13}=sina\)
Pt trở thành:
\(cos2x.cosa+sin2x.sina=1\)
\(\Leftrightarrow cos\left(2x-a\right)=1\)
\(\Leftrightarrow2x-a=k2\pi\)
\(\Leftrightarrow x=\dfrac{a}{2}+k\pi\)
