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1.
\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)
\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)
\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)
\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)
2.
\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)
\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)
\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)
b) \(2sin^2x-3sinxcosx+cos^2x=0\)
\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)
a/
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow3tan^2x+8tanx+8\sqrt{3}-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k\pi\end{matrix}\right.\)
b/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(tan^2x+2tanx-2=\frac{1}{2}\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^2x+4tanx-5=0\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-5\right)+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sinx+1\right)\left(1-2sin^2x-1\right)=0\)
\(\Leftrightarrow sin^2x\left(sinx+1\right)=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
\(y=sin\left(\frac{\pi}{2}-2x\right)=cos2x\Rightarrow y'=-2sin2x\)
= (sin^2x + cos^2x)^2 - 3sin^4x.cos^2x - 3sin^2x.cos^4x
= 1 - 3/4sin^2 (2x).sin^2x - 3/4sin^2(2x).cos^2x
= 1 - 3/4sin^2(2x)
a/
\(\Leftrightarrow3cos^2x-4sinx.cosx+1-cos^2x=1\)
\(\Leftrightarrow2cos^2x-4sinx.cosx=0\)
\(\Leftrightarrow2cosx\left(cosx-2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\tanx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=arctan\left(\frac{1}{2}\right)+k\pi\end{matrix}\right.\)
b.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(4-3tanx+3tan^2x=1+tan^2x\)
\(\Leftrightarrow2tan^2x-3tanx+3=0\)
Pt vô nghiệm