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\(k\in Z\)
a.
\(cos\left(x-2\right)=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=arccos\left(\dfrac{2}{5}\right)+k2\pi\\x-2=-arccos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+arccos\left(\dfrac{2}{5}\right)+k2\pi\\x=2-arcos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
d.
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3>1\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
f.
\(\Leftrightarrow cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
h.
\(cos\left(3x+10^0\right)=-1\)
\(\Leftrightarrow3x+10^0=180^0+k360^0\)
\(\Leftrightarrow3x=170^0+k360^0\)
\(\Leftrightarrow x=\dfrac{1}{3}.170^0+k120^0\)
j.
\(cos\left[cos\left(x+2\right)\right]=1\)
\(\Leftrightarrow cos\left(x+2\right)=k2\pi\)
Do \(-1\le cos\left(x+2\right)\le1\Rightarrow-1\le k2\pi\le1\)
\(\Rightarrow k=0\)
\(\Rightarrow cos\left(x+2\right)=0\)
\(\Rightarrow x+2=\dfrac{\pi}{2}+n\pi\)
\(\Rightarrow x=-2+\dfrac{\pi}{2}+n\pi\)
\(tanx=-tan\dfrac{\pi}{5}\)
\(\Leftrightarrow tanx=tan\left(-\dfrac{\pi}{5}\right)\)
\(\Leftrightarrow x=-\dfrac{\pi}{5}+k\pi\)
Mình quên mất, nó nằm trong khoảng (π/2; π) nha, mình xin lỗi
3.
\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)
\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)
\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)
\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)
\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)
\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)
4.
\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)
\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)
\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)
\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)
\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)
\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)
\(cos^2x=\dfrac{1}{2}\Leftrightarrow2cos^2x-1=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)