Giải phương trình: a) sin (2x - pi/6) = -1/2; b) sin(x/3 + pi/2) = căn 3/2

Giải SBT Toán 11 Bài tập cuối chương 1

Bài 72 trang 32 SBT Toán 11 Tập 1: Giải phương trình:

a) $\sin \left(2x-\frac{\pi }{6}\right)=-\frac{1}{2}$ ;

b) $\sin \left(\frac{x}{3}+\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2}$ ;

c) $\cos \left(2x+\frac{\pi }{5}\right)=\frac{\sqrt{2}}{2}$ ;

d) $2\cos \frac{x}{2}+\sqrt{3}=0$ ;

e) $\sqrt{3}\tan \left(2x+\frac{\pi }{3}\right)-1=0$ ;

g) cot(3x + π) = – 1.

Lời giải:

a) Do $\sin \left(-\frac{\pi }{6}\right)=-\frac{1}{2}$  nên $\sin \left(2x-\frac{\pi }{6}\right)=-\frac{1}{2}$ $\iff \sin \left(2x-\frac{\pi }{6}\right)=\sin \left(-\frac{\pi }{6}\right)$

b) Do $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$  nên $\sin \left(\frac{x}{3}+\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2}$ $\iff \sin \left(\frac{x}{3}+\frac{\pi }{2}\right)=\sin \frac{\pi }{3}$

c) Do $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$  nên  $\cos \left(2x+\frac{\pi }{5}\right)=\frac{\sqrt{2}}{2}$ $\iff \cos \left(2x+\frac{\pi }{5}\right)=\cos \frac{\pi }{4}$

d) $2\cos \frac{x}{2}+\sqrt{3}=0$

$\iff \cos \frac{x}{2}=-\frac{\sqrt{3}}{2}$

  $\iff \cos \frac{x}{2}=\cos \frac{5\pi }{6}$    (do $\cos \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}$ )

e) $\sqrt{3}\tan \left(2x+\frac{\pi }{3}\right)-1=0$

$\iff \tan \left(2x+\frac{\pi }{3}\right)=\frac{1}{\sqrt{3}}$

$\iff \tan \left(2x+\frac{\pi }{3}\right)=\tan \frac{\pi }{6}$            (do $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ )

$\iff 2x+\frac{\pi }{3}=\frac{\pi }{6}+k\pi \left(k\in \mathbb{Z}\right)$

$\iff 2x=-\frac{\pi }{6}+k\pi \left(k\in \mathbb{Z}\right)$

$\iff x=-\frac{\pi }{12}+k\frac{\pi }{2}\left(k\in \mathbb{Z}\right)$.

g) Do $\cot \left(-\frac{\pi }{4}\right)=-1$  nên cot(3x + π) = – 1 $\iff \cot \left(3x+\pi \right)=\cot \left(-\frac{\pi }{4}\right)$

$\iff 3x+\pi =-\frac{\pi }{4}+k\pi \left(k\in \mathbb{Z}\right)$

$\iff 3x=-\frac{5\pi }{4}+k\pi \left(k\in \mathbb{Z}\right)$

$\iff x=-\frac{5\pi }{12}+k\frac{\pi }{3}\left(k\in \mathbb{Z}\right)$.