1.

\(\Leftrightarrow cos\left(2x+\dfrac{4\pi}{3}\right)=0\)

\(\Leftrightarrow2x+\dfrac{4\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=-\dfrac{5\pi}{6}+k\pi\)

\(\Leftrightarrow x=-\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

b.

\(\Leftrightarrow2+2cos\left(2x+\dfrac{\pi}{3}\right)-3=0\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

c.

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

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

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

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

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

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)

a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)

\(\Rightarrow3x+60=30+2k180\)

\(\Rightarrow3x=2k180-30\)

\(\Leftrightarrow x=120k-10\)

Vậy ...

b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)

Vậy ...

c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)

\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)

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

Vậy ...

d, Ta có : \(\cot\left(2x+\pi\right)=-1\)

\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)

\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)

Vậy ...

a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)

Vậy...

d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)

\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)

\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)

Vậy...

Đặt \(x+\dfrac{\pi}{6}=t\Rightarrow x=t-\dfrac{\pi}{6}\Rightarrow3x=3t-\dfrac{\pi}{2}\)

\(2cost=sin\left(3t-\dfrac{\pi}{2}\right)-cos\left(3t-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow2cost=-cos3t-sin3t\)

\(\Leftrightarrow2cost=3cost-4cos^3t+4sin^3t-3sint\)

\(\Leftrightarrow4sin^3t-3sint+cost-4cos^3t=0\)

\(cost=0\) không phải nghiệm

\(\Rightarrow4tan^3t-3tant\left(1+tan^2t\right)+1+tan^2t-4=0\)

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

\(\Leftrightarrow\left(tant+1\right)\left(tan^2t-3\right)=0\)

\(\Leftrightarrow...\)

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

\(\Leftrightarrow2\left(\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx\right)+2cos\left(x-\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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