a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)

b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx

⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x

⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x

⇔ 4sin2x + (sinx + cosx) . sin2x = 0

⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)

⇔ sin2x = 0

c, 2cos³x = sin3x

⇔ 2cos³x = 3sinx - 4sin³x

⇔ 4sin³x + 2cos³x - 3sinx(sin²x + cos²x) = 0

⇔ sin³x + 2cos³x - 3sinx.cos²x = 0

Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình

Xét cosx ≠ 0 chia cả 2 vế cho cos³x ta được : 

tan³x + 2 - 3tanx = 0

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

d, cos²x - \(\sqrt{3}sin2x\) = 1 + sin²x

⇔ cos²x - sin²x - \(\sqrt{3}sin2x\) = 1

⇔ cos2x - \(\sqrt{3}sin2x\) = 1

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

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

e, cos³x + sin³x = 2cos⁵x + 2sin⁵x

⇔ cos³x (1 - 2cos²x) + sin³x (1 - 2sin²x) = 0

⇔ cos³x . (- cos2x) + sin³x . cos2x = 0

⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

chỗ kia là \(2cos\left(\frac{3x}{2}\right)\)nha mn

cos 5x-sin(3x+pi)=0

=>sin(3x+pi)=cos5x

=>sin(3x+pi)=sin(pi/2-5x)

=>3x+pi=pi/2-5x+k2pi hoặc 3x+pi=pi/2+5x+k2pi

=>8x=-pi/2+k2pi hoặc -2x=-pi/2+k2pi

=>x=-pi/16+kpi/4 hoặc x=pi/4-kpi

\(A=cosx+2cos^2x+8cos^3x-6cosx\)

\(=cosx\left(8cos^2x+2cosx-5\right)\)

\(B=\left(sin^3x-2sin^5x\right)-\left(2cos^5x-cos^3x\right)\)

\(=sin^3x\left(1-2sin^2x\right)-cos^3x\left(2cos^2x-1\right)\)

\(=sin^3x.cos2x-cos^3x.cos2x\)

\(=cos2x\left(sin^3x-cos^3x\right)\)

\(=cos2x\left(sinx-cosx\right)\left(1+sinx.cosx\right)\)

1.

\(\Leftrightarrow\sqrt{3}cos^2x-\sqrt{3}+cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)

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

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

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\left[sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

2/

\(\Leftrightarrow3sinx-4sin^3x-\sqrt{3}cosx=2sinx\)

\(\Leftrightarrow4sin^3x-sinx+\sqrt{3}cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow4tan^3x-tanx\left(1+tan^2x\right)+\sqrt{3}\left(1+tan^2x\right)=0\)

\(\Leftrightarrow3tan^3x+\sqrt{3}tan^2x-tanx+\sqrt{3}=0\)

Bạn xem lại đề, pt bậc 3 này ko giải được (nghiệm rất xấu)

a/

\(\Leftrightarrow sinx.cosx\left(sin^2x-cos^2x\right)=\frac{\sqrt{2}}{8}\)

\(\Leftrightarrow2sinx.cosx\left(cos^2x-sin^2x\right)=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin2x.cos2x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow\frac{1}{2}sin4x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin4x=-\frac{\sqrt{2}}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{16}+\frac{k\pi}{2}\\x=\frac{5\pi}{16}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

Câu này đề hơi kì quái, bạn coi lại đề được ko? Biến đổi mấy cách vẫn thấy ko ổn

c/

\(\Leftrightarrow\left(2sinx-cosx+1\right)\left(1+cosx\right)=1-cos^2x\)

\(\Leftrightarrow\left(2sinx-cosx+1\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\)

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

\(\left(1\right)\Leftrightarrow cosx=-1\Leftrightarrow\pi x=\pi+k2\pi\)

\(\left(2\right)\Leftrightarrow2sinx=0\Rightarrow sinx=0\)

\(\Rightarrow x=k\pi\)

Kết hợp lại ta được \(x=k\pi\)

d/

\(\Leftrightarrow2sin8x.cosx=cos\left(\frac{\pi}{2}-2x\right)+1-1-cos\left(\frac{\pi}{2}+4x\right)\) (hạ bậc vế phải)

\(\Leftrightarrow2sin8x.cosx=sin2x+sin4x\)

\(\Leftrightarrow2sin8x.cosx=2sin3x.cosx\)

\(\Leftrightarrow cosx\left(sin8x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin8x=sin3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=3x+k2\pi\\8x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp

b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)

\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)

\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)

\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)

c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:

\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)

Đặt \(\sqrt{tanx+1}=t\ge0\)

\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)

\(\Leftrightarrow3t^3-5t^2+3t-10=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)

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

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

Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)

\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)

\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)