Bộ 8 Đề kiểm tra giữa học kì 1 Toán 11 có đáp án (Mới nhất) - Đề số 8

$\mathrm{x}=\arctan (-\frac{1}{2})+\mathrm{k\pi };\mathrm{x}=\frac{\mathrm{\pi }}{2}+\mathrm{k\pi }(\mathrm{k}\in \mathrm{\mathbb{Z}})$

$\mathrm{x}=\frac{\mathrm{\pi }}{4}+\mathrm{k\pi };\mathrm{x}=\arctan (-2)+\mathrm{k\pi }(\mathrm{k}\in \mathrm{\mathbb{Z}})$

$\mathrm{x}=\arctan 2+\mathrm{k\pi };\mathrm{x}=\frac{\mathrm{\pi }}{4}+\mathrm{k\pi }(\mathrm{k}\in \mathrm{\mathbb{Z}})$

$\mathrm{x}=\arctan 2+\mathrm{k\pi };\mathrm{x}=\frac{\mathrm{\pi }}{3}+\mathrm{k\pi }(\mathrm{k}\in \mathrm{\mathbb{Z}})$

5

7

4

6

cosx.cos2x.cos5x=0

sinx.sin2x.sin4x=0

sinx.sin2x.sin5x=0

cosx.cos2x.cos4x=0

sin (a+b) = sina + sinb

cos(-2a) = 2cosa

$\sin a.\sin b=\frac{1}{2}\left[\sin \right(a-b)-\sin (a+b\left)\right]$

$\cos a-\cos b=-2\sin \frac{a+b}{2}.\sin \frac{a-b}{2}$

$\sin x=\frac{2}{3}$

2sinx - 3cosx = 4

tanx = 2017

$\sin 2x=\frac{1}{3}$

${(\mathrm{x}+1)}^2+{(\mathrm{y}-2)}^2=9$

${(\mathrm{x}-2)}^2+{(\mathrm{y}+1)}^2=9$

${(\mathrm{x}+2)}^2+{(\mathrm{y}+1)}^2=-9$

${(\mathrm{x}-2)}^2+{(\mathrm{y}-1)}^2=9$

$\mathrm{x}\ne \frac{\mathrm{\pi }}{2}+\mathrm{k\pi },\mathrm{k}\in \mathrm{\mathbb{Z}}$

$\mathrm{x}\ne \mathrm{k}2\mathrm{\pi },\mathrm{k}\in \mathrm{\mathbb{Z}}$

$\mathrm{x}\ne \frac{\mathrm{\pi }}{4}+\mathrm{k\pi },\mathrm{k}\in \mathrm{\mathbb{Z}}$

$\mathrm{x}\ne \mathrm{k\pi },\mathrm{k}\in \mathrm{\mathbb{Z}}$

$\sin x=\sin \alpha \iff x=\alpha +k2\mathrm{\pi }\mathrm{hoặc}\mathrm{x}=\mathrm{\pi }-\mathrm{\alpha }+\mathrm{k}2\mathrm{\pi }$

$\mathrm{cox}=\mathrm{cos\alpha }\iff \mathrm{x}=\mathrm{\alpha }+\mathrm{k}2\mathrm{\pi }\mathrm{hoặc}\mathrm{x}=\mathrm{\pi }-\mathrm{\alpha }+\mathrm{k}2\mathrm{\pi }$

$\tan x=\tan \alpha \iff x=\alpha +k\mathrm{\pi }$

$\mathrm{cox}=\mathrm{cos\alpha }\iff \mathrm{x}=\mathrm{\alpha }+\mathrm{k}2\mathrm{\pi }\mathrm{hoặc}\mathrm{x}=-\mathrm{\alpha }+\mathrm{k}2\mathrm{\pi }$

$\mathrm{ℝ}\backslash \left\{-\frac{\mathrm{\pi }}{3}+k2\mathrm{\pi };\mathrm{k}\in \mathrm{\mathbb{Z}}\right\}$

$\mathrm{ℝ}\backslash \left\{-\frac{\mathrm{\pi }}{3}+k\mathrm{\pi };\mathrm{k}\in \mathrm{\mathbb{Z}}\right\}$

$\mathrm{ℝ}\backslash \left\{\frac{\mathrm{\pi }}{6}+k2\mathrm{\pi };\mathrm{k}\in \mathrm{\mathbb{Z}}\right\}$

$\mathrm{ℝ}\backslash \left\{\frac{\mathrm{\pi }}{6}+k\mathrm{\pi };\mathrm{k}\in \mathrm{\mathbb{Z}}\right\}$