a/

\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)

\(\Leftrightarrow3+5cosx=\left(1-cos^2x\right)-cos^2x\)

\(\Leftrightarrow2cos^2x+5cosx+2=0\)

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

\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)

b/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0\)

\(\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0\)

\(a+b+c=\sqrt{3}-\left(\sqrt{3}+1\right)+1=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow6\left(\frac{1-cos2x}{2}\right)+2\left(1-cos^22x\right)=5\)

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

\(\Leftrightarrow cos2x\left(2cos2x+3\right)=0\)

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

\(\Rightarrow2x=\frac{\pi}{2}+k\pi\)

\(\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

d/

\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)

\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^22x+sin2x+1=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)

1.

\(4\left(1-cos^23x\right)+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}-4=0\)

Đặt \(cos3x=a\Rightarrow\left|a\right|\le1\)

\(\Rightarrow4\left(1-a^2\right)+2\left(\sqrt{3}+1\right)a-\sqrt{3}-4=0\)

\(\Leftrightarrow-4a^2+2\left(\sqrt{3}+1\right)a-\sqrt{3}=0\)

\(\Delta'=\left(\sqrt{3}+1\right)^2-4\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\frac{-\sqrt{3}-1+\sqrt{3}-1}{-4}=\frac{1}{2}\\a=\frac{-\sqrt{3}-1-\sqrt{3}+1}{-4}=\frac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cos3x=\frac{1}{2}\\cos3x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\pm\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

2.

\(\Leftrightarrow2cos^2x-1+9cosx+5=0\)

\(\Leftrightarrow2cos^2x+9cosx+4=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cosx=cos\left(\frac{2\pi}{3}\right)\)

\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)

3.

\(\Leftrightarrow4cos^5\left(2-6x\right)+8\left[cos\left(2-6x\right)+1\right]-13=0\)

Đặt \(cos\left(2-6x\right)=a\Rightarrow\left|a\right|\le1\)

\(\Rightarrow4a^5+8a-5=0\)

Bạn coi lại đề bài, pt bậc 5 ko nhẩm được nghiệm thì làm sao mà giải?

4.

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow1+tan^2x-\left(3+\sqrt{3}\right)tanx-3+\sqrt{3}=0\)

\(\Leftrightarrow tan^2x-\left(3+\sqrt{3}\right)tanx-2+\sqrt{3}=0\)

\(\Delta=\left(3+\sqrt{3}\right)^2-4\left(-2+\sqrt{3}\right)=20+2\sqrt{3}\)

Chắc bạn lại nhầm hệ số nào đó, kết quả xấu kinh dị

\(\Rightarrow\left[{}\begin{matrix}tanx=\frac{3+\sqrt{3}-\sqrt{20+2\sqrt{3}}}{2}=tana\\tanx=\frac{3+\sqrt{3}+\sqrt{20+2\sqrt{3}}}{2}=tanb\end{matrix}\right.\)

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

a/

\(\Leftrightarrow cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}sin\left(\frac{3x}{2}+\frac{\pi}{10}\right)\)

Đặt \(\frac{x}{2}+\frac{\pi}{5}=a\Rightarrow\frac{x}{2}=a-\frac{\pi}{5}\Rightarrow\frac{3x}{2}=3a-\frac{3\pi}{5}\)

Pt trở thành:

\(cosa=\frac{1}{2}sin\left(3a-\frac{3\pi}{5}+\frac{\pi}{10}\right)\)

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

\(\Leftrightarrow cosa=-\frac{1}{2}sin\left(\frac{\pi}{2}-3a\right)=-\frac{1}{2}cos3a\)

\(\Leftrightarrow cos3a+2cosa=0\)

\(\Leftrightarrow4cos^3a-3cosa+2cosa=0\)

\(\Leftrightarrow4cos^3a-cosa=0\)

\(\Leftrightarrow cosa\left(4cos^2a-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosa=0\\cosa=\frac{1}{2}\\cosa=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=0\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}+\frac{\pi}{5}=\frac{\pi}{2}+k\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{\pi}{3}+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\) (5 nghiệm bạn tự biến đổi)

b/

ĐKXĐ: ...

Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)

Pt trở thành:

\(4\left(a^2-2\right)+4a=7\)

\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)

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

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

c/

ĐKXĐ: ...

Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)

Pt trở thành:

\(9a+2\left(a^2-4\right)=1\)

\(\Leftrightarrow2a^2+9a-9=0\)

Pt này nghiệm xấu quá bạn :(

d/ĐKXĐ: ...

Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)

Pt trở thành:

\(2\left(a^2+4\right)+9a-1=0\)

\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

a.

\(\Leftrightarrow2sin\frac{17\pi}{30}cos\left(3x-\frac{7\pi}{30}\right)=\sqrt{3}\)

\(\Leftrightarrow cos\left(3x-\frac{7\pi}{30}\right)=\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}\)

Đặt \(\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cos\left(3x-\frac{7\pi}{30}\right)=cosa\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{7\pi}{30}=a+k2\pi\\3x-\frac{7\pi}{30}=-a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{90}+\frac{a}{3}+\frac{k2\pi}{3}\\x=\frac{7\pi}{30}-\frac{a}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)

Chắc bạn ghi sai đề, con số \(\frac{4\pi}{3}\) sẽ hợp lý hơn con số \(\frac{4\pi}{5}\) rất nhiều

b.

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow2tanx.cosx-tanx+1-2cosx=0\)

\(\Leftrightarrow tanx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right)\left(tanx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\tanx=1\end{matrix}\right.\)

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

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)

1.

\(\Leftrightarrow3\left(2cos^22x-1\right)-\left(1-cos^22x\right)+cos2x-2=0\)

\(\Leftrightarrow7cos^22x+cos2x-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\frac{6}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{1}{2}arccos\left(\frac{6}{7}\right)+k\pi\end{matrix}\right.\)

2.

ĐKXĐ: ...

\(\Leftrightarrow1+cot^2x+3cotx+1=0\)

\(\Leftrightarrow cot^2x+3cotx+2=0\)

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

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

3.

ĐKXĐ; ..

\(\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0\)

\(\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.\)

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

4.

\(\Leftrightarrow2cos^2x-1-3cosx=2+2cosx\)

\(\Leftrightarrow2cos^2x-5cosx-3=0\)

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

\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

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

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

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

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

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