Bài tự luận · Bài 13
Dạng 3. Phương trình đưa về dạng phương trình lượng giác cơ bản Phương pháp: - $\sin f(x)=\sin g(x)\Leftrightarrow \left[ \begin{array}{l} f(x)=g(x)+k2\pi \\ f(x)=\pi -g(x)+k2\pi \\ \end{array} (k\in \mathbb{Z}) \right.$. - $\cos f(x)=\cos g(x)\Leftrightarrow \left[ \begin{array}{l} f(x)=g(x)+k2\pi \\ f(x)=-g(x)+k2\pi \\ \end{array} (k\in \mathbb{Z}) \right.$.
Bài tự luận Phương trình lượng giác cơ bản · Bài 13
(CD11) Giải phương trình:
a) $\sin\left( 2x+\frac{\pi}{4} \right)=\sin x$
b) $\sin 2x=\cos 3x$;
c) $\cos^{2}2x=\cos^{2}\left( x+\frac{\pi}{6} \right)$
a) $\sin\left( 2x+\frac{\pi}{4} \right)=\sin x$
b) $\sin 2x=\cos 3x$;
c) $\cos^{2}2x=\cos^{2}\left( x+\frac{\pi}{6} \right)$
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Lời giải
a) $\sin\left( 2x+\frac{\pi}{4} \right)=\sin x\Leftrightarrow \left[ \begin{array}{l} 2x+\frac{\pi}{4}=x+k2\pi \\ 2x+\frac{\pi}{4}=\pi -x+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-\frac{\pi}{4}+k2\pi \\ x=\frac{\pi}{4}+k\frac{2\pi}{3} \end{array} \right.(k\in \mathbb{Z})$.
b) $\sin 2x=\cos 3x$
$\Leftrightarrow \sin 2x=\sin\left( \frac{\pi}{2}-3x \right)\Leftrightarrow \left[ \begin{array}{l} 2x=\frac{\pi}{2}-3x+k2\pi \\ 2x=\pi -\left( \frac{\pi}{2}-3x \right)+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{10}+k\frac{2\pi}{5} \\ x=-\frac{\pi}{2}+k2\pi \end{array} \right.(k\in \mathbb{Z}).$
c) $\cos^{2}2x=\cos^{2}\left( x+\frac{\pi}{6} \right)\Leftrightarrow \frac{1+\cos 4x}{2}=\frac{1+\cos\left( 2x+\frac{\pi}{3} \right)}{2}\Leftrightarrow \cos 4x=\cos\left( 2x+\frac{\pi}{3} \right)$
$\Leftrightarrow \left[ \begin{array}{l} 4x=2x+\frac{\pi}{3}+k2\pi \\ 4x=-(2x+\frac{\pi}{3})+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{6}+k\pi \\ x=-\frac{\pi}{18}+k\frac{\pi}{3} \end{array} \right.(k\in \mathbb{Z}).$
b) $\sin 2x=\cos 3x$
$\Leftrightarrow \sin 2x=\sin\left( \frac{\pi}{2}-3x \right)\Leftrightarrow \left[ \begin{array}{l} 2x=\frac{\pi}{2}-3x+k2\pi \\ 2x=\pi -\left( \frac{\pi}{2}-3x \right)+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{10}+k\frac{2\pi}{5} \\ x=-\frac{\pi}{2}+k2\pi \end{array} \right.(k\in \mathbb{Z}).$
c) $\cos^{2}2x=\cos^{2}\left( x+\frac{\pi}{6} \right)\Leftrightarrow \frac{1+\cos 4x}{2}=\frac{1+\cos\left( 2x+\frac{\pi}{3} \right)}{2}\Leftrightarrow \cos 4x=\cos\left( 2x+\frac{\pi}{3} \right)$
$\Leftrightarrow \left[ \begin{array}{l} 4x=2x+\frac{\pi}{3}+k2\pi \\ 4x=-(2x+\frac{\pi}{3})+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{6}+k\pi \\ x=-\frac{\pi}{18}+k\frac{\pi}{3} \end{array} \right.(k\in \mathbb{Z}).$