2.

\(sin3x+cos2x=1+2sinx.cos2x\)

\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)

\(\Leftrightarrow cos2x+sinx-1=0\)

\(\Leftrightarrow-2sin^2x+sinx=0\)

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

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

1.

\(cos3x-cos4x+cos5x=0\)

\(\Leftrightarrow cos3x+cos5x-cos4x=0\)

\(\Leftrightarrow2cos4x.cosx-cos4x=0\)

\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)

3.

\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)

Đến đây dễ rồi tự làm tiếp nha.

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, (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

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

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

\(\Leftrightarrow2^{\cos2x-1}\left(2\cos x-1\right)=2\cos^2x\left(2\cos x-1\right)\)

\(\Leftrightarrow\left(2\cos x-1\right)\left(2^{\cos2x}-2\cos^2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\cos x=\frac{1}{2}\\2^{\cos2x}=\cos2x+1\end{array}\right.\)

* Với \(\cos x=\frac{1}{2}\) ta có \(x=\frac{\pi}{3}=k2\pi,k\in Z\)

* Với \(2^{\cos2x}=\cos2x+1\) (*), đặt \(t=\cos2x;t\in\left[-1;1\right]\)

Phương trình trở thành \(2^t-t-1=0\)

Xét hàm số \(f\left(t\right)=2^t-t-1,t\in\left[-1;1\right]\)

Có \(f'\left(t\right)=2^t\ln2-1,t\in\left[-1;1\right];f'\left(t\right)=0\) có đúng 1 nghiệm  nên phương trình \(f\left(t\right)=0\) có tối đa 2 nghiệm. Mà \(f\left(0\right)=f\left(1\right)=0\) nên \(t=0;t=1\) là tất cả các nghiệm của phương trình \(f\left(t\right)=0\)

Do đó phương trình (*) \(\Leftrightarrow\left[\begin{array}{nghiempt}\cos2x=0\\\cos2x=1\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+k\frac{\pi}{2}\\x=k\pi\end{array}\right.\) \(k\in Z\)

Vậy phương trình đã cho có 3 nghiệm là :

\(x=\frac{\pi}{3}+k2\pi;x=\frac{\pi}{4}+k\frac{\pi}{2};x=k\pi;k\in Z\)

a/

\(\Leftrightarrow1-2\left(2cos^2x-1\right)-\sqrt{3}sinx+cosx=0\)

\(\Leftrightarrow3-4cos^2x+cosx-\sqrt{3}sinx=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(4cosx+3\right)-\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin^2\frac{x}{2}\left(4cosx+3\right)-2\sqrt{3}sin\frac{x}{2}cos\frac{x}{2}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin\frac{x}{2}=0\Rightarrow x=k2\pi\\sin\frac{x}{2}\left(4cosx+3\right)-\sqrt{3}cos\frac{x}{2}=0\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow sin\frac{x}{2}\left(8cos^2\frac{x}{2}-1\right)-\sqrt{3}cos\frac{x}{2}=0\)

- Với \(\left\{{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-1\end{matrix}\right.\) \(\Rightarrow x=-\pi+k4\pi\) là 1 nghiệm

- Với \(cos\frac{x}{2}\ne0\) chia 2 vế cho \(cos^3\frac{x}{2}\)

\(tan\frac{x}{2}\left(8-1-tan^2\frac{x}{2}\right)-\sqrt{3}-\sqrt{3}tan^2\frac{x}{2}=0\)

\(\Leftrightarrow-tan^3\frac{x}{2}-\sqrt{3}tan^2\frac{x}{2}+7tan\frac{x}{2}-\sqrt{3}=0\)

Đặt \(tan\frac{x}{2}=t\)

\(\Rightarrow t^3+\sqrt{3}t^2-7t+\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\sqrt{3}\\t=-2-\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k\pi\\\frac{x}{2}=-\frac{5\pi}{12}+k\pi\\\frac{x}{2}=\frac{\pi}{12}+k\pi\end{matrix}\right.\)

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

b/

\(\Leftrightarrow cos^2x-sin^2x+cos^2x-sinx.cosx=8\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)+cosx\left(cosx-sinx\right)=8\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(2cosx+sinx-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\left(1\right)\\2cosx+sinx=8\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x-\frac{\pi}{4}=k\pi\)

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

Xét (2), theo điều kiện có nghiệm của pt lượng giác bậc nhất, \(2^2+1^2< 8^2\Rightarrow\left(2\right)\) vô nghiệm

c/

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)

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

Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)

Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cosx.cosa+sinx.sina=cosa\)

\(\Leftrightarrow cos\left(x-a\right)=cosa\)

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

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

\(cos2x+2cos5x.cos2x=0\)

\(\Leftrightarrow cos2x\left(1+2cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos5x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)