Đề là \(sin\left(5-\dfrac{\pi}{3}\right)\) hay \(sin\left(5x-\dfrac{\pi}{3}\right)\) nhỉ?

\(sin3x-cos3x=\sqrt{2}sin\left(5x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=\sqrt{2}sin\left(5x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{3}\right)=sin\left(3x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{\pi}{3}=3x-\dfrac{\pi}{4}+k2\pi\\5x-\dfrac{\pi}{3}=\dfrac{5\pi}{4}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\)

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

Xem lại đề bài đi

Đề sai nhiều chỗ vậy, lần sau ghi đúng đề đi.

\(cos3x+sin7x=2sin^2\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)+2cos^2\dfrac{9x}{2}\)

\(\Leftrightarrow cos3x+sin7x=cos\left(\dfrac{\pi}{2}-5x\right)+1-2cos^2\dfrac{9x}{2}\)

\(\Leftrightarrow cos3x+sin7x=sin5x-cos9x\)

\(\Leftrightarrow2cos6x.cos3x+2cos6x.sinx=0\)

\(\Leftrightarrow2cos6x.\left(cos3x+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+sinx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+cos\left(\dfrac{\pi}{2}-x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\2cos\left(\dfrac{\pi}{4}+x\right).cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos\left(\dfrac{\pi}{4}+x\right)=0\\cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{4}+x=\dfrac{\pi}{2}+k\pi\\2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}\\x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{3\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

ĐKXĐ:...

Biến đổi đoạn trong ngoặc trước cho đỡ rối:

\(cos4x+sin2x=cos\left(3x+x\right)+sin\left(3x-x\right)\)

\(=cos3x.cosx-sin3x.sinx+sin3x.cosx-cos3x.sinx\)

\(=cosx\left(cos3x+sin3x\right)-sinx\left(cos3x+sin3x\right)\)

\(=\left(cosx-sinx\right)\left(cos3x+sin3x\right)\)

Thay vào phương trình:

\(\left(cosx-sinx\right)^2=2\left(sinx+cosx\right)+3\)

\(\Leftrightarrow1-2sinx.cosx=2\left(sinx+cosx\right)+3\)

Đặt \(sinx+cosx=a\Rightarrow-2sinx.cosx=1-a^2\)

\(2-a^2=2a+3\Rightarrow a=-1\Rightarrow sinx+cosx=-1\Rightarrow...\)

\(\Leftrightarrow3sinx-4sin^3x+4cos^3x-3cosx+2cosx=0\)

\(\Leftrightarrow3sinx-cosx-4sin^3x+4cos^3x=0\)

Với \(cosx=0\) ko phải nghiệm, với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow3tanx\left(1+tan^2x\right)-\left(1+tan^2x\right)-4tan^3x+4=0\)

\(\Leftrightarrow-tan^3x-tan^2x+3tanx+3=0\)

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

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

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

Tới đây chắc bạn hoàn thành được phần còn lại

Tham khảo

⇔cosx.sinπ4−sinx.cosπ4=1√2⇔cos⁡�.sin⁡�4−sin⁡�.cos⁡�4=12

⇔⎡⎢⎣π4−x=π4−2kπ⇒x=2kππ4−x=π−π4−2kπ⇒x=−π2+2kπ

a: sin x=-6/5=-1,2

mà -1<=sin x<=1

nên \(x\in\varnothing\)
b: sin3x=căn 3/2

=>3x=pi/3+k2pi hoặc 3x=2/3pi+k2pi

=>x=pi/9+k2pi/3 hoặc x=2/9pi+k2pi/3

c: \(sin\left(x+\dfrac{pi}{3}\right)=sin\left(\dfrac{3}{4}pi\right)\)

=>x+pi/3=3/4pi+k2pi hoặc x+pi/3=1/4pi+k2pi

=>x=5/12pi+k2pi hoặc x=-1/12pi+k2pi

d: =>sin(x+5/6pi)=5/4

mà sin(x+5/6pi) thuộc [-1;1]

nên \(x\in\varnothing\)