Đáp án C

+ y = 2sin²x + 4cos²x + 6sinxcosx = 3sin 2x + cos 2x + 3

+ Hàm số y = 3sin 2x tuần hoàn với chu kì 2π/2 = π.

+ Hàm số y = cos 2x tuần hoàn với chu kì 2π/2 = π.

+ Do đó hàm số y = 2sin²x + 4cos²x + 6sinxcosx là hàm tuần hoàn với chu kì π

\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)

b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)

c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)

\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)

d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)

\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)

e/

1.

\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)

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

Pt trở thành:

\(2\left(1-t^2\right)+3t=0\)

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

\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

2.

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

Pt trở thành:

\(1-t^2-4t=4\)

\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1\)

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

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

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

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

3.

\(\Leftrightarrow1+cosx+sinx+sinx.cosx=2\)

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

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

Pt trở thành:

\(2t+t^2-1-2=0\)

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

\(\Leftrightarrow sinx+cosx=1\)

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

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

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

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

1.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+k\pi\\x=-\frac{7\pi}{24}+k\pi\end{matrix}\right.\)

2.

\(2\left(1-cosx\right)-3\sqrt{3}sinx-\left(1+cosx\right)=4\)

\(\Leftrightarrow cosx+\sqrt{3}sinx=-1\)

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

\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=-\frac{1}{2}\)

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

3.

\(4sinx.cosx-2sinx+1-2cosx=0\)

\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)

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

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

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

4.

\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)

Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)

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

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

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)

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

\(\Leftrightarrow...\)

6.

\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)

\(\Leftrightarrow14sin^2x-5sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)