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 \(x\in\varnothing\)

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

a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)

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

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

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

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

b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)

c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)

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

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)

=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)

\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)

\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)

\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)

\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x\right)cos\left(\dfrac{\pi}{2}\right)-sin\left(x\right)sin\left(\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)-sin\left(x\right)\)

\(=0\)

1.

Chắc đề là \(sin\left[\pi sin2x\right]=1?\)

\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)

Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)

\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)

Thế vào (1)

\(\Rightarrow sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)

Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)

Thế vào (2):

\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\) chắc bạn tự giải tiếp được

3.

\(\Leftrightarrow2sin\left(x+84^0\right).cos\left(60^0\right)=cos20^0\)

\(\Leftrightarrow sin\left(x+84^0\right)=sin70^0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+84^0=70^0+k360^0\\x+84^0=110^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-14^0+k360^0\\x=26^0+k360^0\end{matrix}\right.\)