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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

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

\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)

2.\(sin^22x+cos^23x=1\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)

\(\Leftrightarrow cos6x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)

Vậy...

3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)

\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)

\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))

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

Vậy...

4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)

\(\Leftrightarrow cos2x+cos4x=1+cos6x\)

\(\Leftrightarrow2cos3x.cosx=2cos^23x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

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

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

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

\(a,\sin2x=\dfrac{-1}{2}\Leftrightarrow\sin2x=\sin\left(-\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

a, \(sin2x=-\dfrac{1}{2}\)

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

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

b, \(sin\dfrac{x}{2}=-1\)

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

\(\Leftrightarrow x=-\pi+k2\pi\)

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.

1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)

⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)

⇔ 2cos²x - 5cosx + 2 = 0

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

⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên

2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)

⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0 

⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)

⇒ sin⁴x + cos⁴x = 48.sin⁴x . cos⁴x

⇔ (sin²x + cos²x)² - 2sin²x. cos²x = 3 . (2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)² = 3(2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin⁴2x

⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)

⇔ 1 - 2sin²2x = 0

⇔ cos4x = 0

⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)

⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)

⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)

⇔ sin2x - sin²2x - (1 + cos4x) = 0

⇔ sin2x - sin²2x - 2cos²2x = 0

⇔ sin2x - 2 (cos²2x + sin²2x) + sin²2x = 0

⇔ sin²2x + sin2x - 2 = 0

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

⇔ sin2x = 1

⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

4, cos5x + cos2x + 2sin3x . sin2x = 0

⇔ cos5x + cos2x + cosx - cos5x = 0

⇔ cos2x + cosx = 0

⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)

⇔ \(cos\dfrac{3x}{2}=0\)

⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)

⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)

Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)

⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}

Vậy các nghiệm thỏa mãn là các phần tử của tập hợp 

\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)

5, \(\dfrac{cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx}{sin2x-1}=1\)

⇒ \(cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx=sin2x-1\)

⇒ cos²x + 3sin²x + 3\(\sqrt{2}\)sin2x + 1 = 0

⇔ 2 + 2sin²x + 3\(\sqrt{2}\)sin2x = 0

⇔ 2 + 1 - cos2x + 3\(\sqrt{2}\) sin2x = 0

⇔ \(3\sqrt{2}sin2x-cos2x=-1\)

Còn lại tự giải

7, \(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)

⇔ \(2cos2x.cos\dfrac{\pi}{4}+4sinx=2+\sqrt{2}\left(1-sinx\right)\)

⇔ \(\sqrt{2}cos2x+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)

Dùng công thức : cos2x = 1 - 2sin²x đưa về phương trình bậc 2 ẩn sinx

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