Question

3cotx-2=0

3tanx+4=0

6sinx-3=0

4cotx+2=0

cotx-1=0

sin(x+ \(\frac{\pi}{2}\))=1

sin 3x=0

Answer 1

ĐKXĐ: ...

a/ \(cotx=\frac{2}{3}\Leftrightarrow x=arccot\left(\frac{2}{3}\right)+k\pi\)

b/ \(tanx=-\frac{4}{3}\Leftrightarrow x=arctan\left(-\frac{4}{3}\right)+k\pi\)

c/ \(sinx=\frac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

d/ \(cotx=-\frac{1}{2}\Leftrightarrow x=arccot\left(-\frac{1}{2}\right)+k\pi\)

e/ \(cotx=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

f/ \(sin\left(x+\frac{\pi}{2}\right)=1\Leftrightarrow x+\frac{\pi}{2}=\frac{\pi}{2}+k2\pi\Leftrightarrow x=k2\pi\)

g/ \(3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)