Câu hỏi của Julian Edward

Question

giải các pt

a) $tanx+tan\left(\frac{2\pi}{3}-3x\right)=0$

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

c) $\frac{tan2x-2}{2tan2x+1}=3$

d) $\frac{sinx+\sqrt{3}cosx}{3sinx-\sqrt{3}cosx}=1$

Nguyễn Việt Lâm · Accepted Answer

a/

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

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

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

$\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}$

b/

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

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

$\Rightarrow x=15^0+k180^0$Câu trả lời: a/

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

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

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

$\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}$

b/

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

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

$\Rightarrow x=15^0+k180^0$

Nguyễn Việt Lâm · Answer

c/

ĐKXĐ: ...

$\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)$

$\Leftrightarrow5tan2x=-5$

$\Rightarrow tan2x=-1$

$\Rightarrow2x=-\frac{\pi}{4}+k\pi$

$\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}$

d/

ĐKXĐ: ...

$\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx$

$\Leftrightarrow2sinx=2\sqrt{3}cosx$

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

ĐKXĐ: ...

$\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)$

$\Leftrightarrow5tan2x=-5$

$\Rightarrow tan2x=-1$

$\Rightarrow2x=-\frac{\pi}{4}+k\pi$

$\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}$

d/

ĐKXĐ: ...

$\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx$

$\Leftrightarrow2sinx=2\sqrt{3}cosx$

$\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi$

Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC