Giải bài 6 trang 232 SBT đại số và giải tích 11

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LG a

$\sin 2x = {\cos ^4}\frac{x}{2} - {\sin ^4}\frac{x}{2}$

Lời giải chi tiết:

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

LG b

$3\sin 5x - 2\cos 5x = 3$

Lời giải chi tiết:

$\begin{array}{l}3\sin 5x - 2\cos 5x = 3\\ \Leftrightarrow \frac{3}{{\sqrt {13} }}\sin 5x - \frac{2}{{\sqrt {13} }}\cos 5x = \frac{3}{{\sqrt {13} }}\end{array}$

Đặt $\left\{ \begin{array}{l}\cos \alpha = \frac{3}{{\sqrt {13} }}\\\sin \alpha = \frac{2}{{\sqrt {13} }}\end{array} \right.$ ta có:

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

LG c

$\cos \left( {\frac{\pi }{2} + 5x} \right) + \sin x = 2\cos 3x$

Lời giải chi tiết:

$\begin{array}{l}\cos \left( {\frac{\pi }{2} + 5x} \right) + \sin x = 2\cos 3x\\ \Leftrightarrow - \sin 5x + \sin x = 2\cos 3x\\ \Leftrightarrow \sin x - \sin 5x = 2\cos 3x\\ \Leftrightarrow 2\cos 3x\sin \left( { - 2x} \right) = 2\cos 3x\\ \Leftrightarrow - 2\cos 3x\sin 2x - 2\cos 3x = 0\\ \Leftrightarrow - 2\cos 3x\left( {\sin 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 3x = 0\\\sin 2x = - 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} + k\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + \frac{{k\pi }}{3}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.,k \in \mathbb{Z}\end{array}$

LG d

$\sin 2z + \cos 2z = \sqrt 2 \sin 3z$

Lời giải chi tiết:

$\begin{array}{l}\sin 2z + \cos 2z = \sqrt 2 \sin 3z\\ \Leftrightarrow \sqrt 2 \sin \left( {2z + \frac{\pi }{4}} \right) = \sqrt 2 \sin 3z\\ \Leftrightarrow \sin \left( {2z + \frac{\pi }{4}} \right) = \sin 3z\\ \Leftrightarrow \left[ \begin{array}{l}2z + \frac{\pi }{4} = 3z + k2\pi \\2z + \frac{\pi }{4} = \pi - 3z + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - z = - \frac{\pi }{4} + k2\pi \\5z = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} - k2\pi \\z = \frac{{3\pi }}{{20}} + \frac{{k2\pi }}{5}\end{array} \right.,k \in \mathbb{Z}\end{array}$

SBT Toán lớp 11