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\(sin^4\left(x\right)+cos^4\left(x\right)+2sin^2\left(x\right)cos^2\left(x\right)=\left[sin^2\left(x\right)+cos^2\left(x\right)\right]^2=1^2=1\\ \Rightarrow sin^4\left(x\right)+cos^4\left(x\right)=1-2sin^2\left(x\right)cos^2\left(x\right)\left(đpcm\right)\)
sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2*sin^2x*cos^2x
=1-2*cos^2x*sin^2x
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a: \(\Leftrightarrow2\cdot\sin3x\cdot\cos x-2\cos^2x=0\)
\(\Leftrightarrow\cos x\left(\sin3x-\cos x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\\sin3x=\cos x=\sin\left(\dfrac{\Pi}{2}-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\3x=\dfrac{\Pi}{2}-x+k2\Pi\\3x=\dfrac{\Pi}{2}+x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
b: \(\Leftrightarrow\sin x+\sin5x+\sin^2x=0\)
\(\Leftrightarrow\sin x=0\)
hay \(x=k\Pi\)
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\(\Leftrightarrow8sinx.cosx\left(cos^2x-sin^2x\right)-\left(1+cos4x\right)=1\)
\(\Leftrightarrow4sin2x.cos2x-cos4x=2\)
\(\Leftrightarrow2sin4x-cos4x=2\)
\(\Leftrightarrow\frac{2}{\sqrt{5}}sin4x-\frac{1}{\sqrt{5}}cos4x=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow sin\left(4x+\alpha\right)=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow...\)
Nghiệm xấu quá, bạn tự giải nốt
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\(sina+sinb+sinc+3=0\)
\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)
Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)
\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)
b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)
\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)
\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)
\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2cosx-\left(2cos^2x-1\right)+3=0\)
\(\Leftrightarrow-2cos^2x+2cosx+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2>1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pi+k2\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án D
Dùng công thức cos a.cos b+ sin a. sin b= cos (a-b) để biến đổi phương trình không chứa
α
về dạng giống phương trình có chứa
α
Ta có
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\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)-1\)
\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=cos\left(\dfrac{\pi}{2}-x\right)\)
\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sin\dfrac{x}{2}-cos\dfrac{x}{2}.sinx=1\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}.cos^2\dfrac{x}{2}=1\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\left(1-sin^2\dfrac{x}{2}\right)=1\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow\left(sin\dfrac{x}{2}-1\right)\left(2sin^2\dfrac{x}{2}+2sin\dfrac{x}{2}+1\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow...\)