1.

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\frac{1}{2}sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin2x=2\left(vn\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

2.

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

\(\Leftrightarrow cos2x=cos\left(x-\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow\sqrt{3}cosx-3sinx=2sin5x-2sinx\)

\(\Leftrightarrow\sqrt{3}cosx-sinx=2sin5x\)

\(\Leftrightarrow-\left(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\right)=sin5x\)

\(\Leftrightarrow sin5x=-sin\left(x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{3}-x\right)\)

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

\(\Leftrightarrow...\)

Đây nè:

Câu hỏi của Julian Edward - Toán lớp 11 | Học trực tuyến

Câu hỏi của Julian Edward - Toán lớp 11 | Học trực tuyến

d/

Đặt \(sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\) \(\Rightarrow\left|t\right|\le\sqrt{2}\)

\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\frac{1-t^2}{2}\)

Pt trở thành:

\(6t-1=\frac{1-t^2}{2}\)

\(\Leftrightarrow t^2+12t-3=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{39}-6\\t=-\sqrt{39}-6< -\sqrt{2}\left(l\right)\end{matrix}\right.\) (ủa giáo viên ra đề ngẫu nhiên à?)

\(\Rightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{39}-6}{\sqrt{2}}\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=arcsin\left(\frac{\sqrt{39}-6}{\sqrt{2}}\right)+k2\pi\\x-\frac{\pi}{4}=\pi-arcsin\left(\frac{\sqrt{39}-6}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)

Đáp án C

Học bài trước rồi à :D

a: A=(sinx+cosx)^2-1=m^2-1

b: B=căn (sinx+cosx)^2-4sinxcosx=căn m^2-4(m^2-1)=căn -3m^2+4

c: C=(sin^2x+cos^2x)^2-2(sinx*cosx)^2=1-2m^2

d/ ĐKXĐ: ...

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cos^2x+sin^2x+sinx.cosx\right)}{2cosx+3sinx}=cos^2x-sin^2x\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(1+sinx.cosx\right)}{2cosx+3sinx}=\left(cosx-sinx\right)\left(cosx+sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\\\frac{1+sinx.cosx}{2cosx+3sinx}=sinx+cosx\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow1+sinx.cosx=\left(sinx+cosx\right)\left(2cosx+3sinx\right)\)

\(\Leftrightarrow1+sinx.cosx=2sin^2x+3cos^2x+5sinx.cosx\)

\(\Leftrightarrow2sin^2x+3cos^2x+4sinx.cosx-1=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(2tan^2x+3+4tanx-1-tan^2x=0\)

\(\Leftrightarrow tan^2x+4tanx+2=0\)

\(\Leftrightarrow tanx=-2\pm\sqrt{2}\)

\(\Rightarrow x=arctan\left(-2\pm\sqrt{2}\right)+k\pi\)

c/

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\sinx+4cosx-4=0\left(2\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)

Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cosx.cosa+sinx.sina=cosa\)

\(\Leftrightarrow cos\left(x-a\right)=cosa\)

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

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

Câu 1: Có \(-\dfrac{\pi}{3}\le\)\(x\le\dfrac{\pi}{2}\)

\(\Leftrightarrow\dfrac{1}{2}\le cosx\le1\)

\(\Rightarrow-2\ge-4cosx\ge-4\)

\(\Leftrightarrow\sqrt{3}\ge\sqrt{5-4cosx}\ge1\)

Vậy \(y_{min}=1\)

Câu 2: \(\left(\sqrt{3}+1\right)cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx-\sqrt{3}=0\)

\(\Leftrightarrow cos^2x+\sqrt{3}cos^2x+\sqrt{3}sinx.cosx-sinx.cosx+sinx-cosx-\sqrt{3}=0\)

\(\Leftrightarrow-\sqrt{3}\left(1-cos^2x\right)+\sqrt{3}sinx.cosx+cosx\left(cosx-sinx\right)-\left(cosx-sinx\right)=0\)

\(\Leftrightarrow-\sqrt{3}sin^2x+\sqrt{3}sinx.cosx+\left(cosx-1\right)\left(cosx-sinx\right)=0\)

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

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

\(\Leftrightarrow-\sqrt{2}.sin\left(x-\dfrac{\pi}{4}\right)\left[2sin\left(x+\dfrac{\pi}{6}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\left(1\right)\\sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)

Từ (1) \(\Rightarrow x-\dfrac{\pi}{4}=k\pi\left(k\in Z\right)\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

mà \(x\in\left[0;2\pi\right]\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}\\x=\dfrac{5\pi}{4}\end{matrix}\right.\)

Từ (2)\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

mà \(x\in\left[0;2\pi\right]\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\pi\\x=\dfrac{2\pi}{3}\end{matrix}\right.\)

(Chắc là tìm tổng T?)\(\Rightarrow T=\dfrac{\pi}{4}+\dfrac{5\pi}{4}+0+2\pi+\dfrac{2\pi}{3}=\dfrac{25\pi}{6}\)

Câu 3:

\(f\left(x\right)=\sqrt{sin^2x-4cosx+2m}\)

Để hàm số f(x) có tập xác định là R \(\Leftrightarrow sin^2x-4cosx+2m\ge0\forall x\)

\(\Leftrightarrow-cos^2x-4cosx+1+2m\ge0;\forall x\)

\(\Leftrightarrow2m\ge cos^2x+4cosx-1;\forall x\) (*)

Đặt \(g\left(x\right)=cos^2x+4cosx-1\)

Từ (*) \(\Leftrightarrow2m\ge\max\limits_{x\in R}g\left(x\right)\)

Vẽ bảng biến thiên của g(x) với \(-1\le cosx\le1\) sẽ tìm được max \(g\left(x\right)=4\)

\(\Leftrightarrow2m\ge4\)

\(\Leftrightarrow m\ge2\)

Vậy... (Xem hộ đáp án đúng ko?)

d/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(\Leftrightarrow2\sqrt{2}\left(tanx+1\right)=\frac{3}{cos^2x}+2\)

\(\Leftrightarrow2\sqrt{2}tanx+2\sqrt{2}=3\left(1+tan^2x\right)+2\)

\(\Leftrightarrow3tan^2x-2\sqrt{2}tanx+5-2\sqrt{2}=0\)

Pt vô nghiệm

c/

\(\Leftrightarrow1-sin^2x+\sqrt{3}sinx.cosx-1=0\)

\(\Leftrightarrow\sqrt{3}sinx.cosx-sin^2x=0\)

\(\Leftrightarrow sinx\left(\sqrt{3}cosx-sinx\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=\sqrt{3}\end{matrix}\right.\)

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