Giải bài 1.63 trang 30 sách bài tập toán 11 - Kết nối tri thức với cuộc sống

Đề bài

Giải các phương trình sau:

a) \(\sin 5x + \cos 5x =  - 1\);                      

b) \(\cos 3x - \cos 5x = \sin x\);

c) \(2{\cos ^2}x + \cos 2x = 2\);                           

d) \({\sin ^4}x + {\cos ^4}x = \frac{1}{2}{\sin ^2}2x\).

Lời giải chi tiết

a) \(\sin 5x + \cos 5x =  - 1 \Leftrightarrow \sqrt 2 \sin \left( {5x + \frac{\pi }{4}} \right) =  - 1 \Leftrightarrow \sin \left( {5x + \frac{\pi }{4}} \right) = \frac{{ - 1}}{{\sqrt 2 }}\)                              

\( \Leftrightarrow \left[ \begin{array}{l}5x + \frac{\pi }{4} = \frac{{ - \pi }}{4} + k2\pi \\5x + \frac{\pi }{4} = \pi  + \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ - \pi }}{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{5} + k\frac{{2\pi }}{5}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(\cos 3x - \cos 5x = \sin x \Leftrightarrow 2\sin 4x\sin x = \sin x \Leftrightarrow \sin x\left( {2\sin 4x - 1} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\2\sin 4x - 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\\sin 4x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\4x = \frac{\pi }{6} + k2\pi \\4x = \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{\pi }{{24}} + k\frac{\pi }{2}\\4x = \frac{{5\pi }}{{24}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

c) \(2{\cos ^2}x + \cos 2x = 2 \Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) + \cos 2x = 1 \Leftrightarrow 2\cos 2x = 1 \Leftrightarrow \cos 2x = \frac{1}{2}\)\( \Leftrightarrow 2x =  \pm \frac{\pi }{3} + k2\pi  \Leftrightarrow x =  \pm \frac{\pi }{6} + k\pi \left( {k \in \mathbb{Z}} \right)\)            

d) \({\sin ^4}x + {\cos ^4}x = \frac{1}{2}{\sin ^2}2x \Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2{\sin ^2}x{\cos ^2}x = \frac{1}{2}{\sin ^2}2x\)

\( \Leftrightarrow 1 - \frac{1}{2}{\sin ^2}2x - \frac{1}{2}{\sin ^2}2x = 0 \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \)
\( \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)\)