a.

\(cos2x=sinx\)

\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}-x\right)\)

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

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

b.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}sin\frac{x}{2}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{5\pi}{3}+k4\pi\\x=-\frac{5\pi}{3}+k4\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}-\sqrt{3}cos\frac{x}{2}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\frac{2\pi}{3}+k4\pi\\x=\frac{4\pi}{3}+k4\pi\end{matrix}\right.\)

d.

ĐKXĐ: ...

\(\Leftrightarrow tan\left(3x-\frac{\pi}{5}\right)=tan\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow3x-\frac{\pi}{5}=\frac{\pi}{2}-x+k\pi\)

\(\Leftrightarrow x=\frac{7\pi}{40}+\frac{k\pi}{4}\)

e.

ĐKXĐ: \(\left\{{}\begin{matrix}cos3x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{\pi}{6}+\frac{k\pi}{3}\)

\(\frac{sin3x.sinx}{cos3x.cosx}=1\)

\(\Leftrightarrow cos3x.cosx=sin3x.sinx\)

\(\Leftrightarrow cos3x.cosx-sin3x.sinx=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

ĐKXĐ: \(sinx\ne\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{6}+k2\pi\\x\ne\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Pt tương đương:

\(cosx=\sqrt{3}sinx\Leftrightarrow tanx=\frac{1}{\sqrt{3}}\)

\(\Leftrightarrow x=\frac{\pi}{6}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\frac{7\pi}{6}+k2\pi\)

Bạn xem lại đề. Mình thấy $x\in (0; \frac{\pi}{4}]$ thì hợp lý hơn @_@

Nếu miền giá trị của x có "chạm" vào \(\frac{\pi}{4}\) thì:

\(y^2=\left(a.1+b.\sqrt{sinx}+c.\sqrt{cosx}\right)^2\)

\(\Rightarrow y^2\le\left(a^2+b^2+c^2\right)\left(1+sinx+cosx\right)\)

\(\Rightarrow y^2\le3\left[1+\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\right]\le3\left(1+\sqrt{2}\right)\)

\(\Rightarrow y\le\sqrt{3+3\sqrt{2}}\)

\(M=\sqrt{3+3\sqrt{2}}\) khi \(\left\{{}\begin{matrix}x=\frac{\pi}{4}\\b=c=\sqrt{\frac{6-3\sqrt{2}}{2}}\\a=\sqrt{3\sqrt{2}-3}\end{matrix}\right.\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(y^2=(a+b\sqrt{\sin x}+c\sqrt{\cos x})^2\leq (a^2+b^2+c^2)(1+\sin x+\cos x)=3(1+\sin x+\cos x)\)

$(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x=1+\sin 2x\leq 1+1=2$ với mọi $x\in (0;\frac{\pi}{4}]$

$\Rightarrow \sin x+\cos x\leq \sqrt{2}$

$\Rightarrow 1+\sin x+\cos x\leq \sqrt{2}+1$

Do đó: $y^2\leq 3(1+\sqrt{2})$

$\Rightarrow y\leq \sqrt{3+3\sqrt{2}}$

Vậy $M=\sqrt{3+3\sqrt{2}}$

a/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow3tanx-\sqrt{3}=0\)

\(\Rightarrow tanx=\frac{1}{\sqrt{3}}\)

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

b/

ĐKXĐ: \(sinx\ne-1\)

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

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

\(\Leftrightarrow\frac{\sqrt{3}cosx-1}{1+sinx}=-1\) (do 2-sinx>0 với mọi x)

\(\Leftrightarrow\sqrt{3}cosx-1=-1-sinx\)

\(\Leftrightarrow sinx=-\sqrt{3}cosx\Rightarrow tanx=-\sqrt{3}\)

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

c/

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)

\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)

\(\Leftrightarrow1-cosx=sin^2x\)

\(\Leftrightarrow1-cosx=1-cos^2x\)

\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

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

d/

ĐKXĐ: \(cosx\ne0\)

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

\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)

\(\Leftrightarrow tanx=\sqrt{3}\)

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

\( a){\mathop{\rm sinx}\nolimits} + \cos x = \sqrt 2 \sin 5x\\ \Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .\sin 5x\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 5x\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = 5x + k2\pi \\ x + \dfrac{\pi }{4} = \pi - 5x + k2\pi \end{array} \right.\left( {k \in \mathbb {Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)

\( b)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \sin \dfrac{\pi }{2}\cos 2x + \cos \dfrac{\pi }{2}\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + 1.\cos 2x + 0.\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \cos 2x - 1 = 0\\ \Leftrightarrow 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} + 1 - 2{\sin ^2}x - 1 = 0\\ \Leftrightarrow \sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - si{n^2}x = 0\\ \Leftrightarrow {\mathop{\rm sinx}\nolimits} \left( {\sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\mathop{\rm sinx}\nolimits} = 0\\ \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \sin \left( {\dfrac{\pi }{3} - x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \dfrac{\pi }{3} - x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{3} - k\pi \end{array} \right. \)

Nhiều quá @@ Tách ra đi ><

\( d)3{\sin ^2}x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 2{\sin ^2}x + 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - 3 = 0\\ *sinx = 0 \Rightarrow \text{không là nghiệm phương trình}\\ *sin \ne 0\\ 2 + 2\sqrt 3 \cot x - 3\left( {1 + {{\cot }^2}x} \right) = 0\\ \Leftrightarrow 3{\cot ^2}x - 2\sqrt 3 \cot x + 1 = 0\\ \Leftrightarrow \cot x = \dfrac{{\sqrt 3 }}{3} \Rightarrow x = \dfrac{\pi }{3} + k\pi \)