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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)
\(sin\left(x+1\right)=\dfrac{2}{3}\Leftrightarrow\left[{}\begin{matrix}x+1=arcsin\dfrac{2}{3}+k2\pi\\x+1=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\dfrac{2}{3}-1+k2\pi\\x=\pi-arcsin\dfrac{2}{3}-1+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\).
a: \(\Leftrightarrow\tan\left(x-\dfrac{\Pi}{5}\right)=-\cot x=\tan\left(x+\dfrac{\Pi}{2}\right)\)
\(\Leftrightarrow x-\dfrac{\Pi}{5}=x+\dfrac{\Pi}{2}+k\Pi\)
\(\Leftrightarrow k\Pi=-\dfrac{7}{10}\Pi\)
hay k=-7/10(vô lý)
b: \(\Leftrightarrow\cos x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{3}+k2\Pi\\x=-\dfrac{\Pi}{3}+k2\Pi\end{matrix}\right.\)
\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)
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+ k2π ⇔ x = .png)
, (k ∈ Z)..png)
) = 0 ⇔ .png)
, (k ∈ Z)..png)
= sin(-600) nên phương trình đã cho tương đương với.png)
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a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)
b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)
c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)
d) \(x=300^0+k540^0,k\in\mathbb{Z}\)