a) Ta có:
sin(x+1)=23⇔[x+1=arcsin23+k2πx+1=π−arcsin23+k2π⇔[x=−1+arcsin23+k2πx=−1+π−arcsin23+k2π;k∈Zsin(x+1)=23⇔[x+1=arcsin23+k2πx+1=π−arcsin23+k2π⇔[x=−1+arcsin23+k2πx=−1+π−arcsin23+k2π;k∈Z
b) Ta có:
sin22x=12⇔1−cos4x2=12⇔cos4x=0⇔4x=π2+kπ⇔x=π8+kπ4,k∈Zsin22x=12⇔1−cos4x2=12⇔cos4x=0⇔4x=π2+kπ⇔x=π8+kπ4,k∈Z
c) Ta có:
cot2x2=13⇔⎡⎢⎣cotx2=√33(1)cotx2=−√33(2)(1)⇔cotx2=cotπ3⇔x2=π3+kπ⇔x=2π3+k2π,k∈z(2)⇔cotx2=cot(−π3)⇔x2=−π3+kπ⇔x=−2π3+k2π;k∈Zcot2x2=13⇔[cotx2=33(1)cotx2=−33(2)(1)⇔cotx2=cotπ3⇔x2=π3+kπ⇔x=2π3+k2π,k∈z(2)⇔cotx2=cot(−π3)⇔x2=−π3+kπ⇔x=−2π3+k2π;k∈Z
d) Ta có:
tan(π12+12x)=−√3⇔tan(π12+12π)=tan(−π3)⇔π12+12=−π3+kπ⇔x=−5π144+kπ12,k∈Z
Vậy nghiệm của phương trình đã cho là: x=−5π144+kπ12,k∈Z
a)
\(sin\left(x+1\right)=\dfrac{2}{3}\Leftrightarrow\left[{}\begin{matrix}x+1=arcsin\dfrac{2}{3}+k2\pi\\x+1=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\dfrac{2}{3}-1+k2\pi\\x=\pi-arcsin\dfrac{2}{3}-1+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\).
b) \(sin^22x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}sin2x=-\dfrac{\sqrt{2}}{2}\\sin2x=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(sin2x=\dfrac{\sqrt{2}}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{4}+k2\pi\\2x=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+k\pi\\2x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+k\pi\\x=\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
\(sin2x=\dfrac{-\sqrt{2}}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-\pi}{4}+k2\pi\\2x=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\).
Vậy phương trình có 4 hệ nghiệm:
\(x=\dfrac{\pi}{8}+k\pi;x=\dfrac{3\pi}{8}+k\pi;x=-\dfrac{\pi}{8}+k\pi\); \(x=\dfrac{5\pi}{8}+k\pi\).
c) \(cot^2\dfrac{x}{2}=\dfrac{1}{3}\)\(\Leftrightarrow\left[{}\begin{matrix}cot\dfrac{x}{2}=\dfrac{1}{\sqrt{3}}\\cot\dfrac{x}{2}=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(cot\dfrac{x}{2}=\dfrac{1}{\sqrt{3}}\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{3}+k\pi\)\(\Leftrightarrow x=\dfrac{2\pi}{3}+k2\pi\).
\(cot\dfrac{x}{2}=\dfrac{-1}{\sqrt{3}}\Leftrightarrow\dfrac{x}{2}=\dfrac{-\pi}{3}+k\pi\)\(\Leftrightarrow x=\dfrac{-2\pi}{3}+k2\pi\).
\(tan\left(\dfrac{\pi}{12}+12x\right)=-\sqrt{3}\Leftrightarrow\dfrac{\pi}{12}+12x=-\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow12x=\dfrac{-5\pi}{12}+k\pi\)\(\Leftrightarrow x=-\dfrac{15\pi}{144}+\dfrac{k\pi}{12}\).
