Chứng minh các biểu thức đã cho không phụ thuộc vào x.

f(x) = 1 ⇒ f′(x) = 0

\( 2)\sin x + \sin 2x + \sin 3x = 0\\ \Leftrightarrow 2\sin 2x.\cos x + \sin 2x = 0\\ \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\ 2\cos x + 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = k\pi \\ \cos x = \dfrac{{ - 1}}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z} } \right) \)

\( 3)\sin x + \sin 2x + \sin 3x + \sin 4x = 0\\ \Leftrightarrow \left( {\sin x + \sin 4x} \right) + \left( {\sin 2x + \sin 3x} \right) = 0\\ \Leftrightarrow 2\sin \dfrac{{5x}}{2}.\cos \dfrac{{3x}}{2} + 2\sin \dfrac{{5x}}{2}.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.\left( {\cos \dfrac{{3x}}{2} + \cos \dfrac{x}{2}} \right) = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.2\cos x.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin \dfrac{{5x}}{2} = 0\\ 2\cos x = 0\\ \cos \dfrac{x}{2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{2k\pi }}{5}\\ x = \dfrac{\pi }{2} + k\pi \\ x = \pi + 2k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x+\frac{1}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin\left(3x+\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(\frac{1+cos2x}{2}\right)-3\sqrt{3}sin2x-4\left(\frac{1-cos2x}{2}\right)=-4\)

\(\Leftrightarrow3cos2x-3\sqrt{3}sin2x=-3\)

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

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

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

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

c/

Ủa đề câu này bạn ghi đúng ko? Nhìn kì kì, cos8x hay cos3x bên vế phải vậy?

d/

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

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

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

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

e/

\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)

\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)

\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)

a.

\(\Leftrightarrow4cos^32x-3cos2x+3cos2x=\sqrt{2}\)

\(\Leftrightarrow cos^32x=\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow cos2x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{4}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{8}+k\pi\) (\(k\in Z\))

c.

ĐKXĐ: \(\left\{{}\begin{matrix}cos3x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan3x.tanx=1\)

\(\Rightarrow tan3x=\dfrac{1}{tanx}\)

\(\Rightarrow tan3x=cotx\)

\(\Rightarrow tan3x=tan\left(\dfrac{\pi}{2}-x\right)\)

\(\Rightarrow3x=\dfrac{\pi}{2}-x+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

b.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x-\left(\dfrac{1}{2}+\dfrac{1}{2}cos6x\right)=0\)

\(\Leftrightarrow cos6x=-cos2x\)

\(\Leftrightarrow cos6x=cos\left(\pi-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}8x=\pi+k2\pi\\4x=-\pi+k2\pi\end{matrix}\right.\)

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

(\(k\in Z\))

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

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

Nhiều quá @@ Tách ra đi ><

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

\(sin2x=\sqrt{3}cos2x\)

Nhận thấy cos2x=0 ko phải nghiệm, pt tương đương:

\(\frac{sin2x}{cos2x}=\sqrt{3}\Leftrightarrow tan2x=\sqrt{3}\)

\(\Rightarrow2x=\frac{\pi}{3}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

b/

\(cos\left(90^0-x\right)=-sin2x=cos\left(2x+90^0\right)\)

\(\Rightarrow\left[{}\begin{matrix}90^0-x=2x+90^0+k360^0\\90^0-x=-2x-90^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k120^0\\x=-180^0+k360^0\end{matrix}\right.\)

c/ Giống câu a

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

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{5}{\sqrt{3}}>1\left(l\right)\end{matrix}\right.\)

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

b.

\(\Leftrightarrow\frac{1}{2}sin4x.cos4x+\frac{1}{8}=0\)

\(\Leftrightarrow\frac{1}{4}sin8x+\frac{1}{8}=0\)

\(\Leftrightarrow sin8x=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{6}+k2\pi\\8x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=\frac{7\pi}{48}+\frac{k\pi}{4}\end{matrix}\right.\)

c.

\(\Leftrightarrow2sin2x.cos2x+\sqrt{3}sin2x=0\)

\(\Leftrightarrow sin2x\left(2cos2x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{5\pi}{12}+k\pi\\x=-\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-\sqrt{2}< -1\left(l\right)\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

\(sin^2x+cosx.cos3x+sin2x.cos2x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x+\dfrac{1}{2}sin4x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}-sin^2x+\dfrac{1}{2}sin4x+\dfrac{1}{2}cos4x=0\)

\(\Leftrightarrow sin4x+cos4x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(4x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(4x+\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)

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

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

\(\left(sin2x+1\right)\left(\sqrt{3}cosx-sinx-1\right)=0\)

\(\Leftrightarrow\left(sin2x+1\right)\left(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left(sin2x+1\right)\left(cos\left(x+\dfrac{\pi}{6}\right)-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)