Đề kiểm tra 15 phút – Chương 1 – Đề số 5 – Đại số và giải tích 11

 Đề bài

Bài 1: Giải các phương trình lượng giác sau:

a) $\tan \left(2x-1\right)=\sqrt{3}$

b) $\cot \left(3x+9^0\right)=\frac{\sqrt{3}}{3}$

c) $\sin \left(3x+1\right)=\sin \left(x-2\right)$

d) $\cos 3x-\sin 2x=0$

e) $\tan \left(2x+1\right)+\cot x=0$

g) $\sin \left(x-{120}^0\right)+\cos 2x=0$

Bài 2: Tìm điều kiện xác định của hàm số $y=\sqrt{\frac{1+{\cot }^{2}x}{1-\sin 3x}}$

Lời giải chi tiết

Bài 1:

a) $\tan \left(2x-1\right)=\sqrt{3}$(1)

ĐK: $\cos (2x-1)\ne 0$

$\iff 2x-1\ne \frac{\pi }{2}+k\pi$

$\iff x\ne \frac{\pi }{4}+\frac{1}{2}+k\frac{\pi }{2}$

Pt (1) $\iff \tan (2x-1)=\tan \frac{\pi }{3}$

$\iff 2x-1=\frac{\pi }{3}+k\pi$

$\iff x=\frac{\pi }{6}+\frac{1}{2}+k\frac{\pi }{2}(k\in Z)\left(tmdk\right)$

b) $\cot \left(3x+9^0\right)=\frac{\sqrt{3}}{3}$   (2)

ĐK: $\sin (3x+9^0)\ne 0$

$\iff 3x+9^0\ne k{180}^0$

$\iff x\ne -3^0+k{60}^0$

(2) $\iff \cot (3x+9^0)=\cot {60}^0$

$\iff 3x+9^0={60}^0+k{180}^0$

$\iff x={17}^0+k{60}^0(k\in Z)$

c) $\sin \left(3x+1\right)=\sin \left(x-2\right)$

$\iff [\begin{array}{c}3x+1=x-2+k2\pi \\ 3x+1=\pi -x+2+k2\pi \end{array}$

$\iff [\begin{array}{c}x=\frac{-3}{2}+k\pi \\ x=\frac{\pi }{4}+\frac{1}{4}+k\frac{\pi }{2}\end{array}(k\in Z)$

d) $\cos 3x-\sin 2x=0$

$\iff \cos 3x=\sin 2x$

$\iff \cos 3x=\cos (\frac{\pi }{2}-2x)$

$\iff [\begin{array}{c}3x=\frac{\pi }{2}-2x+k2\pi \\ 3x=2x-\frac{\pi }{2}+k2\pi \end{array}$

$\iff [\begin{array}{c}x=\frac{\pi }{10}+k\frac{2\pi }{5}\\ x=-\frac{\pi }{2}+k2\pi \end{array}(k\in Z)$

e) $\tan \left(2x+1\right)+\cot x=0$  (3)

ĐK: $\{\begin{array}{c}\cos (2x+1)\ne 0\\ \sin x\ne 0\end{array}$

$\iff \{\begin{array}{c}2x+1\ne \frac{\pi }{2}+k\pi \\ x\ne k\pi \end{array}$

$\iff \{\begin{array}{c}x\ne \frac{\pi }{4}-\frac{1}{2}+k\frac{\pi }{2}\\ x\ne k\pi \end{array}$

$\left(3\right)\iff \tan (2x+1)=-\cot x$

$\iff \tan (2x+1)=\tan (\frac{\pi }{2}+x)$

$\iff 2x+1=\frac{\pi }{2}+x+k\pi$

$\iff x=\frac{\pi }{2}-1+k\pi (k\in Z)$

g) $\sin \left(x-{120}^0\right)+\cos 2x=0$

$\iff \cos 2x=-\sin (x-{120}^0)$

$\iff \cos 2x=\cos ({90}^0+x-{120}^0)$

$\iff \cos 2x=\cos \left(-{30}^0+x\right)$

$\iff [\begin{array}{c}2x=-{30}^0+x+k{360}^0\\ 2x={30}^0-x+k{360}^0\end{array}$

$\iff [\begin{array}{c}x=-{30}^0+k{360}^0\\ x={10}^0+k{120}^0\end{array}(k\in Z)$

Bài 2: $y=\sqrt{\frac{1+{\cot }^{2}x}{1-\sin 3x}}$

 ĐK: $\{\begin{array}{c}1-\sin 3x\ne 0\\ \sin x\ne 0\\ \frac{1+{\cot }^{2}x}{1-\sin 3x}\ge 0\end{array}$

$\iff \{\begin{array}{c}1-\sin 3x\ne 0\\ \sin x\ne 0\end{array}$

$\iff \{\begin{array}{c}\sin 3x\ne 1\\ \sin x\ne 0\end{array}$

$\iff \{\begin{array}{c}3x\ne \frac{\pi }{2}+k2\pi \\ x\ne k\pi \end{array}$

$\iff \{\begin{array}{c}x\ne \frac{\pi }{6}+k\frac{2\pi }{3}\\ x\ne k\pi \end{array}$