ĐKXĐ: ...

\(\Leftrightarrow\frac{cos\left(x+\frac{5\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}+\frac{sin\left(2x-\frac{\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}=0\)

\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)+sin\left(2x-\frac{\pi}{6}\right)=0\)

\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=-sin\left(2x-\frac{\pi}{6}\right)\)

\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=cos\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=x+\frac{5\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

a/

Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)

Pt trở thành:

\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)

\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)

\(\Leftrightarrow cos^2a+4sina-4=0\)

\(\Leftrightarrow1-sin^2a+4sina-4=0\)

\(\Leftrightarrow-sin^2a+4sina-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)

\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)

b/

Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)

Pt trở thành:

\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)

\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)

\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)

\(\Leftrightarrow10sin^2a+4sina-14=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

c/

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)

ĐKXĐ: ...

\(\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k2\pi\\x=-\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)

\(P=\sin^2x+cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)\)

\(=\sin^2x+cos^2\left(\frac{\pi}{3}\right)-sin^2x\)

\(=\cos^2\left(\frac{\pi}{3}\right)=\frac{1}{4}\)

=> P không phụ thuộc vào x

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

Câu 2 bạn coi lại đề

3.

\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)

\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)

\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

4.

Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm

5.

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)

\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)

\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)

\(\Leftrightarrow2sin^3x-sinx-1=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)

\(\Leftrightarrow...\)

6.

\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)

\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)

\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)

=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)

c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)

=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)

=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)

=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)

=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)

=>\(x=\dfrac{1}{4}\Omega+k\Omega\)

\(\frac{1}{2}.cos\left(x+\frac{\pi}{5}\right)+\frac{\sqrt{3}}{2}.sin\left(x+\frac{\pi}{5}\right)=0\)

\(\Leftrightarrow cos\left(\frac{\pi}{3}\right).cos\left(x+\frac{\pi}{5}\right)+sin\left(\frac{\pi}{3}\right).sin\left(x+\frac{\pi}{5}\right)=0\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{5}-\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow cos\left(x-\frac{2\pi}{15}\right)=0\)

\(\Leftrightarrow x-\frac{2\pi}{15}=\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\frac{19\pi}{30}+k\pi\) \(k\in\left[-2020;-1\right]\)