1.

\(\Leftrightarrow sin5x+\sqrt{3}cos5x=-2sin15x\)

\(\Leftrightarrow\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x=-sin15x\)

\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(-15x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=-15x+k2\pi\\5x+\frac{\pi}{3}=\pi+15x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{60}+\frac{k\pi}{10}\\x=-\frac{\pi}{15}+\frac{k\pi}{5}\end{matrix}\right.\)

2.

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

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

Do \(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)\le1\\sin\left(x+\frac{\pi}{6}\right)\le1\end{matrix}\right.\) với mọi x

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)\le2\)

Đẳng thức xảy ra khi và chỉ khi:

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

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

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

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

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

Có b nào gipus mk với cần gấp gấp :)

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

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

Nhiều quá @@ Tách ra đi ><

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

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/

\(\Leftrightarrow\left[{}\begin{matrix}cos2x+1=0\\cos2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=-2\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow2x=\pi+k2\pi\)

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

b/

\(\Leftrightarrow cos5x=sin40^0\)

\(\Leftrightarrow cos5x=cos50^0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=50^0+k360^0\\5x=-50^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10^0+k72^0\\x=-10^0+k72^0\end{matrix}\right.\)

c/

\(\Leftrightarrow sin3x=-cosx\)

\(\Leftrightarrow sin3x=sin\left(x-\frac{\pi}{2}\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{3\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

d/

\(\Leftrightarrow2sinx.cosx+\sqrt{3}sinx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{\sqrt{3}}{2}=cos\left(\frac{5\pi}{6}\right)\end{matrix}\right.\)

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