Câu hỏi của Julian Edward

Question

giải các pt

a) $tan2x+tan40^o=0$

b) $tan\left(2x-15^0\right)-1=0$

c) $3tan\left(60^o-x\right)+\sqrt{3}=0$

d) $tan\left(3x+\frac{2\pi}{5}\right)+tan\frac{\pi}{5}=0$

Nguyễn Việt Lâm · Accepted Answer

a/

$\Leftrightarrow tan2x=-tan40^0$

$\Leftrightarrow tan2x=tan\left(-40^0\right)$

$\Rightarrow2x=-40^0+k180^0$

$\Rightarrow x=-20^0+k90^0$

b/

$\Leftrightarrow tan\left(2x-15^0\right)=1$

$\Rightarrow2x-15^0=45^0+k180^0$

$\Rightarrow x=30^0+k90^0$Câu trả lời: a/

$\Leftrightarrow tan2x=-tan40^0$

$\Leftrightarrow tan2x=tan\left(-40^0\right)$

$\Rightarrow2x=-40^0+k180^0$

$\Rightarrow x=-20^0+k90^0$

b/

$\Leftrightarrow tan\left(2x-15^0\right)=1$

$\Rightarrow2x-15^0=45^0+k180^0$

$\Rightarrow x=30^0+k90^0$

Nguyễn Việt Lâm · Answer

c/

$\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}$

$\Rightarrow60^0-x=-30^0+k180^0$

$\Rightarrow x=90^0+k180^0$

d/

$\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)$

$\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)$

$\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi$

$\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}$Câu trả lời: c/