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

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

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

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

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

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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

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

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

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

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

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

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

$x\ne k2\pi ,k\in \mathbb{Z}$

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

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

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

$\mathrm{cox}=\mathrm{cos\alpha }\iff x=\alpha +k2\pi \mathrm{hoặc}x=\pi -\alpha +k2\pi$

$\tan x=\tan \alpha \iff x=\alpha +k\pi$

$\mathrm{cox}=\mathrm{cos\alpha }\iff x=\alpha +k2\pi \mathrm{hoặc}x=-\alpha +k2\pi$

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

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

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

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