Lời giải:

ĐKXĐ: ...............

PT \(\Leftrightarrow \frac{(\sin x-\cos x)(\sin ^2x+\sin x\cos x+\cos ^2x)}{\sqrt{\sin x}+\sqrt{\cos x}}=-2(\sin x-\cos x)(\sin x+\cos x)\)

\(\Leftrightarrow (\sin x-\cos x)\left[\frac{\sin ^2x+\sin x\cos x+\cos ^2x}{\sqrt{\sin x}+\sqrt{\cos x}}+2(\sin x+\cos x)\right]=0\)

Dễ thấy với $\sin x, \cos x\geq 0$ thì biểu thức trong ngoặc vuông luôn lớn hơn $0$

Do đó:

$\sin x-\cos x=0$

$\Leftrightarrow \sin x=\cos x$

Mà $\sin ^2x+\cos ^2x=1; \sin x, \cos x\geq 0$ nên $\sin x=\cos x=\frac{1}{\sqrt{2}}$

$\Rightarrow x=k\pi -\frac{7}{4}\pi$ với $k$ nguyên.

a) Ta có : \(sin\left(x-\frac{2\pi}{3}\right)=cos2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{18}+k\frac{2\pi}{3}\\x=-\frac{7\pi}{6}-k2\pi\end{matrix}\right.\)

Vậy ...

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, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)

\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)

b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)

\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)

\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)

\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)

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

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

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