a/

\(\sqrt{3}sin4x-cos4x=2cosx\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)

\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

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

\(\Leftrightarrow2\left(sinx+cosx\right)^3-4sinx.cosx\left(sinx+cosx\right)=\sqrt{2}\)

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

\(\Rightarrow2t^3-2t\left(t^2-1\right)=\sqrt{2}\)

\(\Leftrightarrow2t=\sqrt{2}\Leftrightarrow t=\frac{\sqrt{2}}{2}\)

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

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

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

Đề thế này hả bạn?

\(A=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx\) (1)

Đặt \(\frac{\pi}{2}-x=t\Rightarrow dx=-dt;\left\{{}\begin{matrix}x=0\Rightarrow t=\frac{\pi}{2}\\x=\frac{\pi}{2}\Rightarrow t=0\end{matrix}\right.\)

\(A=\int\limits^0_{\frac{\pi}{2}}\frac{\sqrt{cost}}{\sqrt{cost}+\sqrt{sint}}\left(-dt\right)=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{cost}}{\sqrt{sint}+\sqrt{cost}}dt=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx\) (2)

Cộng vế với vế của (1) và (2):

\(2A=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx+\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx=\int\limits^{\frac{\pi}{2}}_0dx=\frac{\pi}{2}\)

\(\Rightarrow A=\frac{\pi}{4}\)

b/ \(B=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx\)

Từ (2) ta thấy \(B=A=\frac{\pi}{4}\)

Khi rõ ra bạn.

đặt t=sinx+cosx và phải có đk của t

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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