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Question

Giải các pt sau :

$tan^2x+cot^2x=1+cos^2\left(3x+\frac{\pi}{4}\right)$

$cos\left(\frac{2\pi}{3}sinx-\frac{2\pi}{3}\right)=1$

cot$\left[\frac{\pi}{4}\left(cosx-1\right)\right]=-1$

Nguyễn Việt Lâm · Accepted Answer

b.

$\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi$

$\Leftrightarrow sinx-1=3k$

$\Leftrightarrow sinx=3k+1$

Do $-1\le sinx\le1$

$\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0$

$\Rightarrow k=0$

$\Rightarrow sinx=1$

$\Rightarrow x=\frac{\pi}{2}+k2\pi$Câu trả lời: b.

$\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi$

$\Leftrightarrow sinx-1=3k$

$\Leftrightarrow sinx=3k+1$

Do $-1\le sinx\le1$

$\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0$

$\Rightarrow k=0$

$\Rightarrow sinx=1$

$\Rightarrow x=\frac{\pi}{2}+k2\pi$

Nguyễn Việt Lâm · Answer

c.

ĐKXĐ: ...

$\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi$

$\Leftrightarrow cosx-1=4k-1$

$\Leftrightarrow cosx=4k$

Mà $-1\le cosx\le1\Rightarrow-1\le4k\le1$

$\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0$

$\Rightarrow cosx=0$

$\Rightarrow x=\frac{\pi}{2}+k\pi$Câu trả lời: c.