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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) \(x=300^0+k540^0,k\in\mathbb{Z}\)
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: \(\Leftrightarrow\cos x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{3}+k2\Pi\\x=-\dfrac{\Pi}{3}+k2\Pi\end{matrix}\right.\)
Bài tập này áp dụng công thức phụ - chéo:
cot(a)=tan(\(\dfrac{\Pi}{2}\)-a) (cái này chắc bạn không quên chứ hihi)
Điều kiện: cos(2x+\(\dfrac{\Pi}{4}\))\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{8}\)+\(\dfrac{k\Pi}{2}\)
cos(\(\Pi\)-\(\dfrac{x}{2}\))\(\ne\)0<=>x\(\ne\)\(\Pi\)-2\(\Pi\)
PT<=>tan(2x+\(\dfrac{\Pi}{4}\))=\(\dfrac{1}{tan\left(\Pi-\dfrac{x}{2}\right)}\)
<=>tan(2x+\(\dfrac{\Pi}{4}\))=cot(\(\Pi\)-\(\dfrac{x}{2}\))
<=>tan(2x+\(\dfrac{\Pi}{4}\))=tan(\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\))
<=>2x+\(\dfrac{\Pi}{4}\)=\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\)
<=>x=-\(\dfrac{\Pi}{2}\)+k\(\dfrac{2\Pi}{3}\)(k\(\in\)Z)
Chúc bạn học tốt. Thân!
\(\dfrac{\Pi}{4}\)\(\Pi\)\(\Pi\)
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)đề là \(tan\left(x-15^0\right)=\frac{\sqrt{3}}{3}\)
Vì \(\frac{\sqrt{3}}{3}=tan30^0\) nên
\(\Leftrightarrow tan\left(x-15^0\right)=tan30^0\)
\(\Leftrightarrow x-15^0=30^0+k180^0\)
\(\Leftrightarrow x=45^0+k180^0\left(k\in Z\right)\)
Đk:\(sin3x\ne0\) và \(cos\frac{2\pi}{5}\ne0\)
\(\Leftrightarrow\frac{cos3x}{sin3x}-\frac{sin\frac{2\pi}{5}}{cos\frac{2\pi}{5}}=0\)
\(\Leftrightarrow cos3x\cdot cos\frac{2\pi}{5}-sin\frac{2\pi}{5}\cdot sin3x=0\)
\(\Leftrightarrow cos\left(3x+\frac{2\pi}{5}\right)=0\)
\(\Leftrightarrow3x+\frac{2\pi}{5}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{30}+\frac{k\pi}{3}\)
\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)
a: =>x-pi/3=pi/4+kpi
=>x=7/12pi+kpi
b: =>x+48 độ=25 độ+k*180
=>x=-23 độ+k*180 độ
c: =>x+3/4pi=pi/7+kpi
=>x=-17/28pi+kpi