Đáp án:
b. \(\left[ \begin{array}{l}
x = k2\pi \\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\pi .\cos x = \dfrac{\pi }{2} + k2\pi \\
\to \cos x = \dfrac{1}{2} + 2k\left( {k \in Z} \right)\\
Xét:k = 0\\
\to \cos x = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.\dfrac{\pi }{4}.\left( {\cos x - \sin x} \right) = \dfrac{\pi }{4} + k\pi \\
\to \cos x - \sin x = 1 + 4k\left( {k \in Z} \right)\\
Xét:k = 0\\
\to \cos x - \sin x = 1\\
\to \dfrac{1}{{\sqrt 2 }}.\cos x - \dfrac{1}{{\sqrt 2 }}\sin x = \dfrac{1}{{\sqrt 2 }}\\
\to \sin \dfrac{\pi }{4}.\cos x - \cos \dfrac{\pi }{4}.\sin x = \dfrac{1}{{\sqrt 2 }}\\
\to \sin \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{1}{{\sqrt 2 }}\\
\to \left[ \begin{array}{l}
\dfrac{\pi }{4} - x = \dfrac{\pi }{4} + k2\pi \\
\dfrac{\pi }{4} - x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k2\pi \\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
