\(P=4sin^2x+\sqrt{2}\left(sin2x.cos\frac{\pi}{4}+cos2x.sin\frac{\pi}{4}\right)\)

\(P=4sin^2x+sin2x+cos2x\)

\(P=2\left(1-cos2x\right)+sin2x+cos2x\)

\(P=2+sin2x-cos2x\)

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

Do \(sin\left(2x-\frac{\pi}{4}\right)\le1\Rightarrow P\le2+\sqrt{2}\)

\(\Rightarrow P_{max}=2+\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=1\Leftrightarrow x=\frac{3\pi}{8}+k\pi\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

Ta có: \(0\le\sin^22x\le1\)

=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)

=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)

=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)

=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)

Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)

\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)

Ta có: \(0\le\sin^2x\le1\)

=>\(0\le6\sin^2x\le6\)

=>\(0-1\le6\sin^2x-1\le6-1\)

=>-1<=f(x)<=5

f(x) min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

f(x) max=5 khi \(\sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

e/

\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)

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

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

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

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

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

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

d/

\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)

\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)

\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)

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

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

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

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

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

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

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

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left\lbrack\frac12\cdot2\cdot\sin x\cdot cosx\right\rbrack^2=5-2\cdot\left\lbrack\frac12\cdot\sin2x\right\rbrack^2\)

\(=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

\(0\le\sin^22x\le1\)

=>\(0\ge-\frac12\sin^22x\ge-\frac12\)

=>\(0+5\ge-\frac12\sin^22x+5\ge-\frac12+5\)

=>\(5\ge-\frac12\sin^22x+5\ge\frac92\)

=>\(\frac92\le-\frac12\sin^22x+5\le5\)

=>\(\sqrt{\frac92}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{2}{3\sqrt2}\ge\frac{1}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1}{\sqrt5}\)

=>\(\frac{2\cdot4}{3\sqrt2}\ge\frac{1\cdot4}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1\cdot4}{\sqrt5}\)

=>\(\frac{4\sqrt2}{3}\ge y\ge\frac{4}{\sqrt5}\)

=>\(y_{\max}=\frac{4\sqrt2}{3}\) khi \(-\frac12\cdot\sin^22x+5=\frac92\)

=>\(-\frac12\cdot\sin^22x=-\frac12\)

=>\(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4}{\sqrt5}\) khi \(-\frac12\cdot\sin^22x+5=5\)

=>\(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\left(1-cos^2x\right)+5\cdot cos^2x-4\left(2\cdot cos^2x-1\right)-2\)

\(=3-3\cdot cos^2x+5\cdot cos^2x-8\cdot cos^2x+4-2=-6\cdot cos^2x+5\)

Ta có: \(0<=cos^2x\le1\)

=>\(0\ge-6\cdot cos^2x\ge-6\)

=>\(0+5\ge-6\cdot cos^2x+5\ge-6+5\)

=>5>=y>=-1

Do đó: \(y_{\min}=-1\) khi \(-6\cdot cos^2x+5=-1\)

=>\(-6\cdot cos^2x=-6\)

=>\(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

y max=5 khi \(-6\cdot cos^2x+5=5\)

=>\(-6\cdot cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

a, \(y=sin^2x-2sinx+3cos^2x\)

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

\(=3-2sinx-2sin^2x\)

Đặt \(sinx=t\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y=f\left(t\right)=3-2t-2t^2\)

\(\Rightarrow y_{min}=min\left\{f\left(0\right);f\left(1\right)\right\}=-1\)

\(y_{max}=max\left\{f\left(0\right);f\left(1\right)\right\}=3\)

b, \(y=sinx-cosx+sin2x+5\)

\(=sinx-cosx-\left(sinx-cosx\right)^2+6\)

Đặt \(sinx-cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(\Rightarrow y=f\left(t\right)=-t^2+t+6\)

\(\Rightarrow y_{min}=min\left\{f\left(-\sqrt{2}\right);f\left(0\right)\right\}=4-\sqrt{2}\)

\(y_{max}=max\left\{f\left(-\sqrt{2}\right);f\left(0\right)\right\}=6\)

\(S=sinx+siny+sin\left(3x+y\right)-sin\left(3x+y\right)-sin\left(x+y\right)\)

\(=sinx+siny-sin\left(x+y\right)\)

\(S^2=\left(sinx+siny-sin\left(x+y\right)\right)^2\le3\left(sin^2x+sin^2y+sin^2\left(x+y\right)\right)\)

\(S^2\le3\left(1-\dfrac{1}{2}\left(cos2x+cos2y\right)+sin^2\left(x+y\right)\right)\)

\(S^2\le3\left[1-cos\left(x+y\right)cos\left(x-y\right)+1-cos^2\left(x-y\right)\right]\)

\(S^2\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)-\left[cos\left(x-y\right)-\dfrac{1}{2}cos\left(x+y\right)\right]^2\right]\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)\right]\)

\(S^2\le3\left(2+\dfrac{1}{4}\right)=\dfrac{27}{4}\)

\(\Rightarrow S\le\dfrac{3\sqrt{3}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=3\\c=2\end{matrix}\right.\)

Câu 2 bạn coi lại đề

3.

\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)

\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)

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

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

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

\(\Leftrightarrow...\)

4.

Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm

5.

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)

\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)

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

\(\Leftrightarrow2sin^3x-sinx-1=0\)

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

\(\Leftrightarrow...\)

1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)