d la sai

\(\Leftrightarrow1+2sinx.cosx-\left(sinx+cosx\right)=0\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

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

thank bạn nhiều

\(y=\sqrt{3}cos2x+2sinxcosx-2\)

\(=\sqrt{3}cos2x+sin2x-2\)

Ta có: \(\left|\sqrt{3}cos2x+sin2x\right|\le\sqrt{\left(\sqrt{3}\right)^2+1^2}=2\)

Do đó \(-2\le\sqrt{3}cos2x+sin2x\le2\)

\(\Leftrightarrow-4\le\sqrt{3}cos2x+sin2x-2\le2\).

Ta có: \(\left|\sqrt{3}cosx-sinx\right|\le\sqrt{\left(\sqrt{3}\right)^2+\left(-1\right)^2}=2\)

Do đó \(-2\le\sqrt{3}cosx-sinx\le2\)

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) \(\Rightarrow2sinx.cosx=t^2-1\)

Do \(x\in\left[0;\frac{\pi}{2}\right]\Rightarrow x+\frac{\pi}{4}\in\left[\frac{\pi}{4};\frac{3\pi}{4}\right]\) \(\Rightarrow\frac{\sqrt{2}}{2}\le sin\left(x+\frac{\pi}{4}\right)\le1\)

\(\Rightarrow1\le t\le\sqrt{2}\)

Pt trở thành: \(m\left(t+1\right)=t^2\Leftrightarrow m=\frac{t^2}{t+1}\)

Xét \(f\left(t\right)=\frac{t^2}{t+1}\) trên \(\left[1;\sqrt{2}\right]\)

Có \(f\left(t\right)-\frac{1}{2}=\frac{t^2}{t+1}-\frac{1}{2}=\frac{\left(t-1\right)\left(2t+1\right)}{2\left(t+1\right)}\ge0\Rightarrow f\left(t\right)\ge\frac{1}{2}\)

\(f\left(t\right)-2\sqrt{2}+2=\frac{t^2}{t+1}-2\sqrt{2}+2=\frac{\left(t-\sqrt{2}\right)\left(t+2-\sqrt{2}\right)}{t+1}\le0\Rightarrow f\left(t\right)\le2\sqrt{2}-2\)

\(\Rightarrow\frac{1}{2}\le m\le2\sqrt{2}-2\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinxcosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(\left(1+\sqrt{2}\right)t-t^2+1-1-\sqrt{2}=0\)

\(\Leftrightarrow t^2-\left(1+\sqrt{2}\right)t+\sqrt{2}=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow x=?\)

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

Đúng rồi ạ. Mình cảm ơn ạ

c/

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

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

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

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

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

b/

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

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

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

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

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

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

Khi rõ ra bạn.