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\(PT\Leftrightarrow2+\frac{1}{sinxcosx}-cotx=-2-sinx-cosxcotx-tanx\)
\(\Leftrightarrow\frac{1}{sinxcosx}-\frac{cosx}{sinx}=-2sinx-\frac{sinx}{cosx}\)
\(\Leftrightarrow1-cos^2x+2sin^2xcosx+sin^2x=0\)
\(\Leftrightarrow2sin^2x+2sin^2xcosx=0\)
\(\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2k\pi\\x=\pi+2k\pi\end{matrix}\right.\Leftrightarrow x=k\pi\)
Mình sửa lại câu trả lời
ĐK:\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(PT\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loai\right)\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow x=\pi+2k\pi\left(loai\right)\)
Vậy phương trình vô nghiệm
ĐKXĐ: ....
Bạn học cách đặt ẩn phụ \(t=tan\frac{x}{2}\) chưa nhỉ? Rồi thì bài này ngắn, còn chưa thì hơi dài, để an toàn cứ coi như chưa học đi:
Ta có: \(tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2}}=\frac{sinx}{cosx+1}\)
Thay vào pt:
\(cotx+sinx\left(1+\frac{sin^2x}{cosx\left(1+cosx\right)}\right)=4\)
\(\Leftrightarrow cotx+sinx\left(\frac{cosx+cos^2x+sin^2x}{cosx\left(1+cosx\right)}\right)-4=0\)
\(\Leftrightarrow cotx+\frac{sinx}{cosx}-4=0\)
\(\Leftrightarrow\frac{1}{tanx}+tanx-4=0\)
Đặt \(t=tanx\Rightarrow t^2-4t+1=0\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2+\sqrt{3}=tan\left(\frac{5\pi}{12}\right)\\tanx=2-\sqrt{3}=tan\left(\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
b.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{3}cot\pi x=\frac{\pi}{6}+k\pi\)
\(\Leftrightarrow cot\pi x=\frac{1}{2}+3k\)
\(\Leftrightarrow\pi x=arccot\left(\frac{1}{2}+3k\right)+n\pi\)
\(\Leftrightarrow x=\frac{1}{\pi}arccot\left(\frac{1}{2}+3k\right)+n\)
c.
\(\Leftrightarrow\left[{}\begin{matrix}\pi tan3x=\frac{\pi}{6}+k2\pi\\\pi tan3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tan3x=\frac{1}{6}+2k\\tan3x=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}arctan\left(\frac{1}{6}+2k\right)+\frac{n2\pi}{3}\\x=\frac{1}{3}arctan\left(\frac{5}{6}+2k\right)+\frac{n2\pi}{3}\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\pi}{2}sin\pi\left(x+1\right)=\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow sin\pi\left(x+1\right)=\frac{1}{2}+2k\)
Do \(-1\le sin\pi\left(x+1\right)\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi\left(x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{6}+k2\pi\\\pi\left(x+1\right)=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{6}+2k\\x+1=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{6}+2k\\x=-\frac{1}{6}+2k\end{matrix}\right.\)
a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \;\; \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
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)
\(sin\left(x+1\right)=\dfrac{2}{3}\Leftrightarrow\left[{}\begin{matrix}x+1=arcsin\dfrac{2}{3}+k2\pi\\x+1=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\dfrac{2}{3}-1+k2\pi\\x=\pi-arcsin\dfrac{2}{3}-1+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\).
tham khảogiúp mình nhé: (tanx + cotx)^2 - (tanx + cotx) = 2? | Yahoo Hỏi & Đáp