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Đặt \(x+\frac{\pi}{4}=t\Rightarrow x=t-\frac{\pi}{4}\)
Pt trở thành:
\(sin^3t=\sqrt{2}sin\left(t-\frac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint-cost\)
\(\Leftrightarrow sint-sin^3t-cost=0\)
\(\Leftrightarrow sint\left(1-sin^2t\right)-cost=0\)
\(\Leftrightarrow sint.cos^2t-cost=0\)
\(\Leftrightarrow cost\left(sint.cost-1\right)=0\)
\(\Leftrightarrow cost\left(\frac{1}{2}sin2t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=2>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c/
ĐKXĐ: ...
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)=-3\)
\(\Leftrightarrow tan\left(\frac{\pi}{9}-2x\right)=-\sqrt{3}\)
\(\Rightarrow\frac{\pi}{9}-2x=-\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{2\pi}{9}+\frac{k\pi}{2}\)
d/
\(\Leftrightarrow\left[{}\begin{matrix}tanx=5\\tan2x=tan4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\2x=4+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\x=2+\frac{k\pi}{2}\end{matrix}\right.\)
a/
ĐKXĐ: ...
\(\Leftrightarrow tanx-8\sqrt{3}=3tanx-6\sqrt{3}\)
\(\Leftrightarrow2tanx=-2\sqrt{3}\)
\(\Rightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)
b/
\(\Leftrightarrow tan2x=-cot\left(\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{\pi}{2}+\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{9\pi}{8}\right)\)
\(\Rightarrow2x=\frac{9\pi}{8}+k\pi\Rightarrow x=\frac{9\pi}{16}+\frac{k\pi}{2}\)
b/ ĐKXĐ: \(cos2x\ne0\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(6sinx-2cos^3x=\frac{10sin2x.cos2x.sinx}{2cos2x}\)
\(\Leftrightarrow6sinx-2cos^3x=5sin2x.sinx\)
\(\Leftrightarrow3sinx-cos^3x=5cosx.sin^2x\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(3tanx\left(1+tan^2x\right)-1=5tan^2x\)
\(\Leftrightarrow3tan^3x-5tan^2x+3tanx-1=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(3tan^2x-2tanx+1\right)=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\) (ko thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
c/
\(a+b+c=1+\sqrt{3}-1-\sqrt{3}=0\)
\(\Rightarrow\) Pt có 2 nghiệm: \(\left[{}\begin{matrix}tanx=1\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow cot^22x+3.cot2x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cot2x=-1\\cot2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k\pi\\2x=arccot\left(-2\right)+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{1}{2}arccot\left(-2\right)+\frac{k\pi}{2}\end{matrix}\right.\)
a/
\(\Leftrightarrow2cos^2x-1+cosx+1=0\)
\(\Leftrightarrow cosx\left(2cosx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\Leftrightarrow tanx+\frac{1}{tanx}=2\)
\(\Leftrightarrow tan^2x+1=2tanx\)
\(\Leftrightarrow tan^2x-2tanx+1=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)
1.
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
2.
\(sinx-\sqrt{3}cosx=2sin5\text{}x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
c/
ĐKXĐ: ...
\(\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)\)
\(\Leftrightarrow5tan2x=-5\)
\(\Rightarrow tan2x=-1\)
\(\Rightarrow2x=-\frac{\pi}{4}+k\pi\)
\(\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
d/
ĐKXĐ: ...
\(\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx\)
\(\Leftrightarrow2sinx=2\sqrt{3}cosx\)
\(\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)
a/
\(\Leftrightarrow tanx=-tan\left(\frac{2\pi}{3}-3x\right)\)
\(\Leftrightarrow tanx=tan\left(3x-\frac{2\pi}{3}\right)\)
\(\Rightarrow x=3x-\frac{2\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}\)
b/
\(tan\left(2x-15^0\right)=tanx\)
\(\Rightarrow2x-15^0=x+k180^0\)
\(\Rightarrow x=15^0+k180^0\)