Bài tự luận · Bài 1
Dạng 1. Giải phương trình lượng giác cơ bản
Bài tự luận Phương trình lượng giác cơ bản · Bài 1
(CD11) Giải phương trình:
a) $\sin\left( 2x-\frac{\pi}{3} \right)=-\frac{\sqrt{3}}{2}$
b) $\cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{2}$
c) $\tan\left( 3x+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{3}$
d) $\sqrt{3}\cot\left( 2x-\frac{\pi}{6} \right)=-1$
a) $\sin\left( 2x-\frac{\pi}{3} \right)=-\frac{\sqrt{3}}{2}$
b) $\cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{2}$
c) $\tan\left( 3x+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{3}$
d) $\sqrt{3}\cot\left( 2x-\frac{\pi}{6} \right)=-1$
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Lời giải
a) Do $\sin\left( -\frac{\pi}{3} \right)=-\frac{\sqrt{3}}{2}$ nên $\sin\left( 2x-\frac{\pi}{3} \right)=-\frac{\sqrt{3}}{2}\Leftrightarrow \sin\left( 2x-\frac{\pi}{3} \right)=\sin\left( -\frac{\pi}{3} \right)$
$\Leftrightarrow \left[ \begin{array}{l} 2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi \\ 2x-\frac{\pi}{3}=\pi -\left( -\frac{\pi}{3} \right)+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=k\pi \\ x=\frac{5\pi}{6}+k\pi \end{array} \right.(k\in \mathbb{Z}).$
b) Do $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ nên $\cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{2}\Leftrightarrow \cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\cos\frac{\pi}{6}$
$\Leftrightarrow \left[ \begin{array}{l} \frac{x}{2}+\frac{\pi}{4}=\frac{\pi}{6}+k2\pi \\ \frac{x}{2}+\frac{\pi}{4}=-\frac{\pi}{6}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-\frac{\pi}{6}+k4\pi \\ x=-\frac{5\pi}{6}+k4\pi \end{array} \right.(k\in \mathbb{Z})$
c) Do $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ nên $\tan\left( 3x+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{3}\Leftrightarrow \tan\left( 3x+\frac{\pi}{4} \right)=\tan\frac{\pi}{6}$
$\Leftrightarrow 3x+\frac{\pi}{4}=\frac{\pi}{6}+k\pi \Leftrightarrow x=-\frac{\pi}{36}+k\frac{\pi}{3}(k\in \mathbb{Z})$
d) $\sqrt{3}\cot\left( 2x-\frac{\pi}{6} \right)=-1\Leftrightarrow \cot\left( 2x-\frac{\pi}{6} \right)=-\frac{1}{\sqrt{3}}$.
Do $\cot\frac{2\pi}{3}=-\frac{1}{\sqrt{3}}$ nên $\cot\left( 2x-\frac{\pi}{6} \right)=-\frac{1}{\sqrt{3}}\Leftrightarrow \cot\left( 2x-\frac{\pi}{6} \right)=\cot\frac{2\pi}{3}$
$\Leftrightarrow 2x-\frac{\pi}{6}=\frac{2\pi}{3}+k\pi \Leftrightarrow x=\frac{5\pi}{12}+k\frac{\pi}{2}(k\in \mathbb{Z}).$
$\Leftrightarrow \left[ \begin{array}{l} 2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi \\ 2x-\frac{\pi}{3}=\pi -\left( -\frac{\pi}{3} \right)+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=k\pi \\ x=\frac{5\pi}{6}+k\pi \end{array} \right.(k\in \mathbb{Z}).$
b) Do $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ nên $\cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{2}\Leftrightarrow \cos\left( \frac{x}{2}+\frac{\pi}{4} \right)=\cos\frac{\pi}{6}$
$\Leftrightarrow \left[ \begin{array}{l} \frac{x}{2}+\frac{\pi}{4}=\frac{\pi}{6}+k2\pi \\ \frac{x}{2}+\frac{\pi}{4}=-\frac{\pi}{6}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-\frac{\pi}{6}+k4\pi \\ x=-\frac{5\pi}{6}+k4\pi \end{array} \right.(k\in \mathbb{Z})$
c) Do $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ nên $\tan\left( 3x+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{3}\Leftrightarrow \tan\left( 3x+\frac{\pi}{4} \right)=\tan\frac{\pi}{6}$
$\Leftrightarrow 3x+\frac{\pi}{4}=\frac{\pi}{6}+k\pi \Leftrightarrow x=-\frac{\pi}{36}+k\frac{\pi}{3}(k\in \mathbb{Z})$
d) $\sqrt{3}\cot\left( 2x-\frac{\pi}{6} \right)=-1\Leftrightarrow \cot\left( 2x-\frac{\pi}{6} \right)=-\frac{1}{\sqrt{3}}$.
Do $\cot\frac{2\pi}{3}=-\frac{1}{\sqrt{3}}$ nên $\cot\left( 2x-\frac{\pi}{6} \right)=-\frac{1}{\sqrt{3}}\Leftrightarrow \cot\left( 2x-\frac{\pi}{6} \right)=\cot\frac{2\pi}{3}$
$\Leftrightarrow 2x-\frac{\pi}{6}=\frac{2\pi}{3}+k\pi \Leftrightarrow x=\frac{5\pi}{12}+k\frac{\pi}{2}(k\in \mathbb{Z}).$