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

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

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/

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

\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)

\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)

b/

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

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

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

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

c/

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)

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

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

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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

1.

\(y=\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)\Rightarrow\) tập giá trị là \(\left[-\sqrt{2};\sqrt{2}\right]\)

2. ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)

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

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

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

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

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

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow2y+y.cosx=sinx+2cosx+3\)

\(\Leftrightarrow sinx+\left(2-y\right)cosx=2y-3\)

\(\Rightarrow1^2+\left(2-y\right)^2\ge\left(2y-3\right)^2\)

\(\Leftrightarrow3y^2-8y+4\le0\)

\(\Rightarrow\frac{2}{3}\le y\le2\)

4.

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

\(\Rightarrow-2\le y\le2\)

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=\frac{1}{2}sin7x-\frac{\sqrt{3}}{2}cos7x\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow...\)

6.

\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)

\(\Leftrightarrow cos4x=4cos2x+5\)

\(\Leftrightarrow2cos^22x-1=4cos2x+5\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

7.

Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn

8.

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

\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)

a/

\(\Leftrightarrow cosx.\frac{1}{2}-\frac{\sqrt{3}}{2}sinx=cos\left(\frac{\pi}{3}-x\right)\)

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

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

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

\(\Rightarrow x=k\pi\)

b/

\(\Leftrightarrow\sqrt{2}sin\left(5x+\frac{\pi}{4}\right)=\sqrt{2}cos13x\)

\(\Leftrightarrow cos\left(\frac{\pi}{4}-5x\right)=cos13x\)

\(\Leftrightarrow\left[{}\begin{matrix}13x=\frac{\pi}{4}-5x+k2\pi\\13x=-\frac{\pi}{4}+5x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{72}+\frac{k\pi}{9}\\x=-\frac{\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)

c/

Đặt \(3cosx-4sinx-6=t\)

Pt trở thành:

\(t^2+2=-3t\Leftrightarrow t^2+3t+2=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3cosx-4sinx-6=-1\\3cosx-4sinx-6=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx-4sinx=5\\3cosx-4sinx=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx.\frac{3}{5}-sinx.\frac{4}{5}=1\\cosx.\frac{3}{5}-sinx.\frac{4}{5}=\frac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+a\right)=1\\cosx\left(x+a\right)=\frac{4}{5}\end{matrix}\right.\) (với góc \(a\in\left[0;\pi\right]\) sao cho \(cosa=\frac{3}{5}\))

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

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

a) Đk: sinx \(\ne\)0<=>x\(\ne\)k\(\Pi\)

pt<=>\(\sqrt{3}\)(1-cos2x)-cosx=0

<=>\(\sqrt{3}\)[1-(2cos²x-1)]-cosx=0

<=>2\(\sqrt{3}\)-2\(\sqrt{3}\)cos²x-cosx=0

<=>\(\left\{{}\begin{matrix}cosx=\dfrac{\sqrt{3}}{2}\\cosx=-\dfrac{2\sqrt{3}}{3}< -1\left(loai\right)\end{matrix}\right.\)

tới đây bạn tự giải cho quen, chứ chép thì thành ra không hiểu gì thì khổ

b)pt<=>2sin2x+2sin²x=1

<=>2sin2x+2sin²x=sin²x+cos²x

<=>4sinx.cosx+sin²x-cos²x=0

Tới đây là dạng của pt đẳng cấp bậc 2, ta thấy cosx=0 không phải là nghiệm của pt nên ta chia cả hai vế của pt cho cos²x:

pt trở thành:

4tanx+tan²x-1=0

<=>\(\left[{}\begin{matrix}tanx=-2+\sqrt{2}\\tanx=-2-\sqrt{5}\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=arctan\left(-2+\sqrt{5}\right)+k\Pi\\x=arctan\left(-2-\sqrt{5}\right)+k\Pi\end{matrix}\right.\)(k thuộc Z)

Chú ý: arctan tương ứng ''SHIFT tan'' (khi thử nghiệm trong máy tính)

c)Đk: cosx\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{2}\)+kpi

pt<=>cos²x+\(\sqrt{3}\)sin²x=1

<=>1-sin²x+\(\sqrt{3}\)sin²x-1=0

<=>(\(\sqrt{3}\)-1)sin²x=0

<=>sinx=0<=>x=k\(\Pi\)(k thuộc Z)

d)

pt<=>\(\sqrt{3}\)sin7x-cos7x=\(\sqrt{2}\)

Khúc này bạn coi SGK trang 35 người ta giả thích rõ ràng rồi

pt<=>\(\dfrac{\sqrt{3}}{2}\)sin7x-\(\dfrac{1}{2}\)cos7x=\(\dfrac{\sqrt{2}}{2}\)

<=>sin(7x-\(\dfrac{\Pi}{3}\))=\(\dfrac{\sqrt{2}}{2}\)

<=>sin(7x-\(\dfrac{\Pi}{3}\))=sin\(\dfrac{\Pi}{4}\)

Tới đây bạn tự giải nhé, giải ra nghiệm rồi kiểm tra xem nghiệm nào thuộc khoảng ( đề cho) rồi kết luận

Câu d) mình nhầm nhé

<=>sin(7x-\(\dfrac{\Pi}{6}\))=\(\dfrac{\sqrt{2}}{2}\) mới đúng sorry