Đề kiểm tra 15 phút – Chương 1 – Đề số 8 – Đại số và giải tích 11

 Đề bài

Bài 1: Giải các phương trình sau:

a) $\cos 2x+9\cos x+5=0$

b) ${\left(\sin x-\cos x\right)}^2$$-\left(\sqrt{2}+1\right)(\sin x-\cos x)+\sqrt{2}=0$

c) $2{\sin }^{2}x+\sqrt{3}\sin 2x=3$

d) $-\sin x-\cos x=\sqrt{2}\sin 5x$

Bài 2: Giải các phương trình sau:

a) $2{\sin }^{2}x+\left(1-\sqrt{3}\right)\sin x\cos x$$+\left(1-\sqrt{3}\right){\cos }^{2}x=1$

b) $1+2\sin x\cos x=\sin x+2\cos x$

Lời giải chi tiết

Bài 1:

$a)\cos 2x+9\cos x+5=0$

$\iff 2{\cos }^{2}x-1+9\cos x+5=0$

$\iff 2{\cos }^{2}x+9\cos +4=0\left(1\right)$

Đặt: $\cos x=t(-1\le t\le 1)$

Khi đó (1) trở thành: $2t^2+9t+4=0$

$\iff [\begin{array}{c}t=\frac{-1}{2}\left(tm\right)\\ t=-4\left(ktm\right)\end{array}$

Với $t=\frac{-1}{2}\Rightarrow \cos x=\frac{-1}{2}$

$\iff \cos x=\cos \frac{2\pi }{3}\iff x=\pm \frac{2\pi }{3}+k2\pi$

$b){\left(\sin x-\cos x\right)}^2$$-\left(\sqrt{2}+1\right)(\sin x-\cos x)+\sqrt{2}=0\left(2\right)$

Đặt: $\sin x-\cos x=t(-\sqrt{2}\le t\le \sqrt{2})$

Khi đó (2) trở thành: $t^2-(\sqrt{2}+1)t+\sqrt{2}=0$

$\iff [\begin{array}{c}t=\sqrt{2}\left(tm\right)\\ t=1\left(tm\right)\end{array}$

Với $t=\sqrt{2}$

$\begin{array}{c}\Rightarrow \sin x-\cos x=\sqrt{2}\\ \iff \sqrt{2}\sin (x-\frac{\pi }{4})=\sqrt{2}\\ \iff \sin (x-\frac{\pi }{4})=1\\ \iff x-\frac{\pi }{4}=\frac{\pi }{2}+k2\pi \\ \iff x=\frac{3\pi }{4}+k2\pi \end{array}$

Với $t=1$

$\begin{array}{c}\Rightarrow \sin x-\cos x=1\\ \iff \sin (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}\\ \iff \sin (x-\frac{\pi }{4})=\sin \frac{\pi }{4}\\ \iff [\begin{array}{c}x-\frac{\pi }{4}=\frac{\pi }{4}+k2\pi \\ x-\frac{\pi }{4}=\pi -\frac{\pi }{4}+k2\pi \end{array}\\ \iff [\begin{array}{c}x=\frac{\pi }{2}+k2\pi \\ x=\pi +k2\pi \end{array}\end{array}$

$\begin{array}{c}c)2{\sin }^{2}x+\sqrt{3}\sin 2x=3\\ \iff 2\frac{1-2\cos x}{2}+\sqrt{3}\sin 2x=3\\ \iff 1-\cos 2x+\sqrt{3}\sin 2x=3\\ \iff \sqrt{3}\sin 2x-\cos 2x=2\\ \iff \frac{\sqrt{3}}{2}\sin 2x-\frac{1}{2}\cos 2x=1\\ \iff \cos \frac{\pi }{6}\sin 2x-\sin \frac{\pi }{6}\cos 2x=1\\ \iff \sin (2x-\frac{\pi }{6})=1\\ \iff 2x-\frac{\pi }{6}=\frac{\pi }{2}=k2\pi \\ \iff x=\frac{\pi }{3}+k\pi \end{array}$

$\begin{array}{c}d)-\sin x-\cos x=\sqrt{2}\sin 5x\\ \iff \frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=-\sin 5x\\ \iff \cos \frac{\pi }{4}\sin x+\sin \frac{\pi }{4}\cos x=\sin (-5x)\\ \iff \sin (x+\frac{\pi }{4})=\sin (-5x)\\ \iff [\begin{array}{c}x+\frac{\pi }{4}=-5x+k2\pi \\ x+\frac{\pi }{4}=\pi +5x+k2\pi \end{array}\\ \iff [\begin{array}{c}6x=-\frac{\pi }{4}+k2\pi \\ -4x=\frac{3\pi }{4}+k2\pi \end{array}\\ \iff [\begin{array}{c}x=-\frac{\pi }{24}+k\frac{\pi }{3}\\ x=-\frac{3\pi }{16}-k\frac{\pi }{2}\end{array}(k\in Z)\end{array}$

Bài 2:

$a)2{\sin }^{2}x+\left(1-\sqrt{3}\right)\sin x\cos x$$+\left(1-\sqrt{3}\right){\cos }^{2}x=1\left(1\right)$

Với $\cos x=0\iff x=\frac{\pi }{2}+k\pi (k\in Z)$

Thay $x=\frac{\pi }{2}+k\pi$ vào phương trình (1) ta được:  2=1 ( vô lí) $\Rightarrow x=\frac{\pi }{2}+k\pi (k\in Z)$ không là nghiệm của (1).

Với $\cos x\ne 0$

$\iff x\ne \frac{\pi }{2}+k\pi (k\in Z)$

$\begin{array}{c}\left(1\right)\iff 2\frac{{\sin }^{2}x}{{\cos }^{2}x}+(1-\sqrt{3})\frac{\sin x}{\cos x}+(1-\sqrt{3})=\frac{1}{{\cos }^{2}x}\\ \iff 2{\tan }^{2}x-(1-\sqrt{3})\tan x+(1-\sqrt{3})=1+{\tan }^{2}x\\ \iff {\tan }^{2}x+(1-\sqrt{3})\tan x-\sqrt{3}=0\\ \iff [\begin{array}{c}\tan x=\sqrt{3}\\ \tan x=-1\end{array}\\ \iff [\begin{array}{c}\tan x=\tan \frac{\pi }{3}\\ \tan x=\tan (-\frac{\pi }{4})\end{array}\\ \iff [\begin{array}{c}x=\frac{\pi }{3}+k\pi \\ x=\frac{-\pi }{4}+k\pi \end{array}(k\in Z)\end{array}$

Vậy phương trình có nghiệm là: $x=\frac{\pi }{3}+k\pi ;x=\frac{-\pi }{4}+k\pi (k\in Z)$

$\begin{array}{c}b)1+2\sin x\cos x=\sin x+2\cos x\\ \iff 1+2\sin x\cos x-\sin x-2\cos x=0\\ \iff (2\sin x\cos x-\sin x)-(2\cos x-1)=0\\ \iff \sin x(2\cos x-1)-(2\cos x-1)=0\\ \iff (2\cos x-1)(\sin x-1)=0\\ \iff [\begin{array}{c}2\cos x-1=0\\ \sin x-1=0\end{array}[\begin{array}{c}\cos x=\frac{1}{2}\\ \sin x=1\end{array}\\ \iff [\begin{array}{c}\cos x=\cos \frac{\pi }{6}\\ \sin x=\sin \frac{\pi }{2}\end{array}\\ \iff [\begin{array}{c}x=\pm \frac{\pi }{6}+k2\pi \\ x=\frac{\pi }{2}+k2\pi \end{array}(k\in Z)\end{array}$