\(\Leftrightarrow sin^3x+3sin^2x+3sinx+1-cos^3x+sinx-cosx+1=0\)

\(\Leftrightarrow\left(sinx+1\right)^3-cos^3x+sinx-cosx+1=0\)

\(\Leftrightarrow\left(sinx-cosx+1\right)\left[\left(sinx+1\right)^2+cosx\left(sinx+1\right)+cos^2x\right]+sinx-cosx+1=0\)

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

\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx+cosx+sinx.cosx+3\right)=0\)

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

Xét (1):

\(\Leftrightarrow2\left(sinx+1\right)+cosx\left(sinx+1\right)+1=0\)

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

Do \(sinx;cosx\ge-1\Rightarrow\left(cosx+2\right)\left(sinx+1\right)\ge0\)

\(\Rightarrow\left(cosx+2\right)\left(sinx+1\right)+1=0\) vô nghiệm

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 \(a\in\left(0;\pi\right)\)

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

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

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

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

a, \(sin^2x-4sinx+3=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow sinx=1\)

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

b, \(2cos^2-cosx-1=0\)

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

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

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

c, \(3sin^2x-2cosx+2=0\)

\(\Leftrightarrow3-3sin^2x+2cosx-5=0\)

\(\Leftrightarrow3cos^2x+2cosx-5=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(3cosx+5\right)=0\)

\(\Leftrightarrow cosx=1\)

\(\Leftrightarrow x=k2\pi\)

mik lm biếng quá mik chỉ nói cách làm thôi nha bạn

1) chia hai vế cho cos^2(x) \(\sqrt{3}tan^2x+\left(1-\sqrt{3}\right)tanx-1+\left(1-\sqrt{3}\right)\left(1+tan^2x\right)=0\)

đặt t = tanx rr giải thôi =D ( máy 570 thì mode5 3 còn máy 580 thì mode 9 2 2) :)))

2) cx làm cách tương tự chia 2 vế cho cos^2x

3) giữ vế trái bung vế phải ra

\(sin2x-2sin^2x=2-4sin^22x\)

đặt t = sin2x (-1=<t=<1)

4) đẩy sinx cosx qua trái hết

\(sinx\left(sin^2-1\right)-cosx\left(cos^2x+1\right)=0\)

\(sinx\left(-cos^2x\right)-cos\left(cos^2x+1\right)=0\)

\(-cos\left(sinxcosx+cos^2x+1\right)=0\)

cái vế đầu cosx=0 bn bik giả rr mà dễ ẹc à còn vế sau thì chia cho cos^2(x) như mấy bài trên rr sau đó đặt t = tanx rr bấm máy là ra thui :))

5)bung cái hằng đẳng thức ra sau đó đặt t=sinx+cosx (t thuộc [-căn(2) ; căn(2)]

khi đó ta có sinxcosx=1/2 sin2x= 1/2t^2 - 1/2

làm đi là ra à

1.

\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

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

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

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

Xét (1):

Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)

Xét (1):

Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm

3.

\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)

\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)

\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)

Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm

1.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)

2.

Đề bài thiếu, cos?x

Và x thuộc khoảng nào?

3.

\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)

\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)

\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)

4.

\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)

\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)