Giải bài 60 trang 30 sách bài tập toán 11 - Cánh diều

Đề bài

Giải phương trình:

a) \(\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right)\)

b) \(\cos \left( {2x - \frac{\pi }{3}} \right) = \sin \left( {\frac{\pi }{4} - x} \right)\)

c*) \({\sin ^2}\left( {x + \frac{\pi }{4}} \right) = {\sin ^2}\left( {2x + \frac{\pi }{2}} \right)\)

d*) \({\cos ^2}\left( {2x + \frac{\pi }{2}} \right) = {\sin ^2}\left( {x + \frac{\pi }{6}} \right)\)

e) \(\cos x + \sin x = 0\)

g) \(\sin x - \sqrt 3 \cos x = 0\)

Lời giải chi tiết

a) Ta có:

\(\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{4} = x + \frac{\pi }{6} + k2\pi \\3x - \frac{\pi }{4} = \pi - \left( {x + \frac{\pi }{6}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{{12}} + k2\pi \\4x = \frac{{13\pi }}{{12}} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{24}} + k\pi \\x = \frac{{13\pi }}{{48}} + k\frac{\pi }{2}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)

b) Ta có \(\sin \left( {\frac{\pi }{4} - x} \right) = \cos \left( {\frac{\pi }{2} - \frac{\pi }{4} + x} \right) = \cos \left( {\frac{\pi }{4} + x} \right)\). Phương trình trở thành:

\(\cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left( {x + \frac{\pi }{4}} \right) \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = x + \frac{\pi }{4} + k2\pi \\2x - \frac{\pi }{3} = - \left( {x + \frac{\pi }{4}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\3x = \frac{\pi }{{12}} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\x = \frac{\pi }{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)

c) Sử dụng công thức hạ bậc, ta có:

\({\sin ^2}\left( {x + \frac{\pi }{4}} \right) = \frac{{1 - \cos \left[ {2\left( {x + \frac{\pi }{4}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{2}} \right)}}{2}\),

\({\sin ^2}\left( {2x + \frac{\pi }{2}} \right) = \frac{{1 - \cos \left[ {2\left( {2x + \frac{\pi }{2}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {4x + \pi } \right)}}{2}\)

Phương trình trở thành:

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

\( \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{2} = 4x + \pi + k2\pi \\2x + \frac{\pi }{2} = - \left( {4x + \pi } \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2x = \frac{\pi }{2} + k2\pi \\6x = - \frac{{3\pi }}{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k\pi \\x = - \frac{\pi }{4} + k\frac{\pi }{3}\end{array} \right. \Leftrightarrow x = - \frac{\pi }{4} + k\frac{\pi }{3}\)\(\left( {k \in \mathbb{Z}} \right)\)

d) Sử dụng công thức hạ bậc, ta có:

\({\cos ^2}\left( {2x + \frac{\pi }{2}} \right) = \frac{{1 + \cos \left[ {2\left( {2x + \frac{\pi }{2}} \right)} \right]}}{2} = \frac{{1 + \cos \left( {4x + \pi } \right)}}{2}\)

\({\sin ^2}\left( {x + \frac{\pi }{6}} \right) = \frac{{1 - \cos \left[ {2\left( {x + \frac{\pi }{6}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{3}} \right)}}{2}\)

Phương trình trở thành:

\(\frac{{1 + \cos \left( {4x + \pi } \right)}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{3}} \right)}}{2} \Leftrightarrow \cos \left( {4x + \pi } \right) = - \cos \left( {2x + \frac{\pi }{3}} \right)\)

Mặt khác, ta có \( - \cos \left( {2x + \frac{\pi }{3}} \right) = \cos \left( {\pi + 2x + \frac{\pi }{3}} \right) = \cos \left( {2x + \frac{{4\pi }}{3}} \right)\).

Phương trình trở thành:

\(\cos \left( {4x + \pi } \right) = \cos \left( {2x + \frac{{4\pi }}{3}} \right) \Leftrightarrow \left[ \begin{array}{l}4x + \pi = 2x + \frac{{4\pi }}{3} + k2\pi \\4x + \pi = - \left( {2x + \frac{{4\pi }}{3}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{3} + k2\pi \\6x = - \frac{{7\pi }}{3} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = - \frac{{7\pi }}{{18}} + k\frac{\pi }{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = - \frac{\pi }{{18}} + k\frac{\pi }{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)

e) Ta có \(\frac{1}{{\sqrt 2 }}\left( {\sin x + \cos x} \right) = \sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4} = \sin \left( {x + \frac{\pi }{4}} \right)\).

Do đó, \(\cos x + \sin x = 0 \Leftrightarrow \frac{1}{{\sqrt 2 }}\left( {\cos x + \sin x} \right) = 0 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\)

\( \Leftrightarrow x + \frac{\pi }{4} = k\pi \Leftrightarrow x = - \frac{\pi }{4} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)

f) Nếu \(\cos x = 0\) thì \(\sin x = 0\). Điều này là vô lí, do \({\sin ^2}x + {\cos ^2}x = 1\).

Như vậy \(\cos x \ne 0\). Phương trình trở thành:

\(\sin x = \sqrt 3 \cos x \Leftrightarrow \frac{{\sin x}}{{\cos x}} = \sqrt 3 \Leftrightarrow \tan x = \sqrt 3 \)

Ta có \(\tan \frac{\pi }{3} = \sqrt 3 \). Phương trình trở thành \(\tan x = \tan \frac{\pi }{3} \Leftrightarrow x = \frac{\pi }{3} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)