Bài tự luận · Bài 10
Dạng 1. Giải phương trình lượng giác cơ bản
Bài tự luận Phương trình lượng giác cơ bản · Bài 10
(KNTT11) Giải các phương trình sau:
a) $2\sin\left( \frac{x}{3}+15^{\circ} \right)+\sqrt{2}=0$;
b) $\cos\left( 2x+\frac{\pi}{5} \right)=-1$;
c) $3\tan 2x+\sqrt{3}=0$;
d) $\cot(2x-3)=\cot 15^{\circ}$.
a) $2\sin\left( \frac{x}{3}+15^{\circ} \right)+\sqrt{2}=0$;
b) $\cos\left( 2x+\frac{\pi}{5} \right)=-1$;
c) $3\tan 2x+\sqrt{3}=0$;
d) $\cot(2x-3)=\cot 15^{\circ}$.
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Lời giải
a) Ta có
$\begin{array}{l} 2\sin\left( \frac{x}{3}+15^{\circ} \right)+\sqrt{2}=0\Leftrightarrow \sin\left( \frac{x}{3}+15^{\circ} \right)=-\frac{\sqrt{2}}{2} \\ \Leftrightarrow \left[ \begin{array}{l} \frac{x}{3}+15^{\circ}=-45^{\circ}+k360^{\circ} \\ \frac{x}{3}+15^{\circ}=225^{\circ}+k360^{\circ} \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-180^{\circ}+k1080^{\circ} \\ x=630^{\circ}+k1080^{\circ} \end{array} \right.(k\in \mathbb{Z}). \end{array}$
b) Ta có $\cos\left( 2x+\frac{\pi}{5} \right)=-1\Leftrightarrow 2x+\frac{\pi}{5}=\pi +k2\pi \Leftrightarrow x=\frac{2\pi}{5}+k\pi (k\in \mathbb{Z})$.
c) Ta có $3\tan 2x+\sqrt{3}=0\Leftrightarrow \tan 2x=-\frac{\sqrt{3}}{3}\Leftrightarrow \tan 2x=\tan\left( -\frac{\pi}{6} \right)$
$\Leftrightarrow 2x=-\frac{\pi}{6}+k\pi \Leftrightarrow x=-\frac{\pi}{12}+k\frac{\pi}{2}(k\in \mathbb{Z}).$
d) Ta có $\cot(2x-3)=\cot 15^{\circ}$
$\Leftrightarrow \cot(2x-3)=\cot\frac{\pi}{12}\Leftrightarrow 2x-3=\frac{\pi}{12}+k\pi \Leftrightarrow x=\frac{3}{2}+\frac{\pi}{24}+k\frac{\pi}{2}(k\in \mathbb{Z}).$
$\begin{array}{l} 2\sin\left( \frac{x}{3}+15^{\circ} \right)+\sqrt{2}=0\Leftrightarrow \sin\left( \frac{x}{3}+15^{\circ} \right)=-\frac{\sqrt{2}}{2} \\ \Leftrightarrow \left[ \begin{array}{l} \frac{x}{3}+15^{\circ}=-45^{\circ}+k360^{\circ} \\ \frac{x}{3}+15^{\circ}=225^{\circ}+k360^{\circ} \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-180^{\circ}+k1080^{\circ} \\ x=630^{\circ}+k1080^{\circ} \end{array} \right.(k\in \mathbb{Z}). \end{array}$
b) Ta có $\cos\left( 2x+\frac{\pi}{5} \right)=-1\Leftrightarrow 2x+\frac{\pi}{5}=\pi +k2\pi \Leftrightarrow x=\frac{2\pi}{5}+k\pi (k\in \mathbb{Z})$.
c) Ta có $3\tan 2x+\sqrt{3}=0\Leftrightarrow \tan 2x=-\frac{\sqrt{3}}{3}\Leftrightarrow \tan 2x=\tan\left( -\frac{\pi}{6} \right)$
$\Leftrightarrow 2x=-\frac{\pi}{6}+k\pi \Leftrightarrow x=-\frac{\pi}{12}+k\frac{\pi}{2}(k\in \mathbb{Z}).$
d) Ta có $\cot(2x-3)=\cot 15^{\circ}$
$\Leftrightarrow \cot(2x-3)=\cot\frac{\pi}{12}\Leftrightarrow 2x-3=\frac{\pi}{12}+k\pi \Leftrightarrow x=\frac{3}{2}+\frac{\pi}{24}+k\frac{\pi}{2}(k\in \mathbb{Z}).$