- Đặt \(\left\{{}\begin{matrix}\sin^2\dfrac{x}{2}=a\\\sin x+3=b\end{matrix}\right.\)

\(PTTT:a^2-ab+b-1=0\)

\(\Leftrightarrow-b\left(a-1\right)+\left(a-1\right)\left(a+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a-b=-1\end{matrix}\right.\)

- Thay lại vào phương trình ta được :\(\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\sin^2\dfrac{x}{2}-\sin x-3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\dfrac{1-\cos x}{2}-\sin x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\cos x+2\sin x=-3\end{matrix}\right.\)

Thấy : \(-\sqrt{5}\le2\sin x+\cos x\le\sqrt{5}\)

\(\Rightarrow2\sin x+\cos x=-3\left(L\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin\dfrac{x}{2}=1\\\sin\dfrac{x}{2}=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k4\pi\\x=-\pi+k4\pi\end{matrix}\right.\)\(\left(K\in Z\right)\)

Vậy ....

2.1

a.

\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

\(cosx-\sqrt{3}sinx=1\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)

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

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

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

c.

\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)

Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm

d.

\(cosx-sinx=1\)

\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

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

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

\(a)\;sinx = \frac{{\sqrt 3 }}{2}\)

Vì \(sin\frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\) nên \(sinx = \frac{{\sqrt 3 }}{2} \Leftrightarrow sin\frac{\pi }{3} = sin\frac{\pi }{3}\) \( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\\x = \pi  - \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\\x = \frac{{2\pi }}{3} + k2\pi ,k \in \mathbb{Z}\end{array} \right.\)

Vậy phương trình có nghiệm là \(x = \frac{\pi }{3} + k2\pi \) hoặc \(x = \frac{{2\pi }}{3} + k2\pi \)\(,k \in \mathbb{Z}\).

\(\begin{array}{l}b)\;sin(x + {30^o}) = sin(x + {60^o})\\ \Leftrightarrow \left[ \begin{array}{l}x + {30^o} = x + {60^o} + k{360^o},k \in \mathbb{Z}\\x + {30^o} = {180^o} - x - {60^o} + k{360^o},k \in \mathbb{Z}\end{array} \right.\\ \Leftrightarrow x = {45^o} + k{180^o},k \in \mathbb{Z}.\end{array}\)

Vậy phương trình có nghiệm là \(x = {45^o} + k{180^o},k \in \mathbb{Z}\).

a, \(cos\left(x-\dfrac{\pi}{3}\right)-sin\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow cos\left(x-\dfrac{7\pi}{12}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x-\dfrac{7\pi}{12}=\pm\dfrac{\pi}{4}+k2\pi\)

...

b, \(\sqrt{3}sin2x+2cos^2x=2sinx+1\)

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

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=sinx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

a)Đk:\(sinx\ne1\)

Pt\(\Leftrightarrow sin^2x+sinx=-2\left(sinx-1\right)\)

\(\Leftrightarrow sin^2x+3sinx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{-3+\sqrt{17}}{2}\left(tm\right)\\sinx=\dfrac{-3-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arcc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\\x=\pi-arc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\end{matrix}\right.\)(\(k\in Z\))

b)Đk:\(sinx\ne1\)

Pt \(\Leftrightarrow\dfrac{1-2sin^2x+sinx}{sinx-1}+1=0\)

\(\Leftrightarrow\dfrac{-\left(sinx-1\right)\left(2sinx+1\right)}{sinx-1}+1=0\)

\(\Leftrightarrow-\left(2sinx+1\right)+1=0\)

\(\Leftrightarrow sinx=0\) (tm)

\(\Leftrightarrow x=k\pi,k\in Z\)

Vậy...

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 \(x\in\varnothing\)

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

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

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

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

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

1.

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

\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

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

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

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

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

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