3.

Theo điều kiện của pt lượng giác bậc nhất:

\(m^2+\left(3m+1\right)^2\ge\left(1-2m\right)^2\)

\(\Leftrightarrow10m^2+6m+1\ge4m^2-4m+1\)

\(\Leftrightarrow3m^2+5m\ge0\Rightarrow\left[{}\begin{matrix}m\ge0\\m\le-\frac{5}{3}\end{matrix}\right.\)

4.

\(\Leftrightarrow1-sin^2x-\left(m^2-3\right)sinx+2m^2-3=0\)

\(\Leftrightarrow-sin^2x-m^2sinx+2m^2+3sinx-2=0\)

\(\Leftrightarrow\left(-sin^2x+3sinx-2\right)+m^2\left(2-sinx\right)=0\)

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

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

\(\Leftrightarrow sinx=1-m^2\)

\(\Rightarrow-1\le1-m^2\le1\)

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

1.

Bạn xem lại đề, \(sin^2x\left(\frac{x}{2}-\frac{\pi}{4}\right)\) là sao nhỉ?Có cả x trong lẫn ngoài ngoặc?

2.

ĐKXĐ: \(sinx\ne0\)

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

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

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

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

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

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

9.

\(\Leftrightarrow1-cos2x+m.sin2x=2m\)

\(\Leftrightarrow m.sin2x-cos2x=2m-1\)

Pt đã cho vô nghiệm khi và chỉ khi:

\(m^2+1< \left(2m-1\right)^2\)

\(\Leftrightarrow3m^2-4m>0\Rightarrow\left[{}\begin{matrix}m< 0\\m>\frac{4}{3}\end{matrix}\right.\)

10.

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

\(\Leftrightarrow2sinx.cos\frac{\pi}{4}=0\)

\(\Leftrightarrow sinx=0\)

\(\Leftrightarrow x=k\pi\)

\(\Rightarrow x=\left\{\pi;2\pi;3\pi\right\}\Rightarrow\sum x=6\pi\)

7.

\(sin3x-cosx=0\)

\(\Leftrightarrow sin3x=cosx\)

\(\Leftrightarrow sin3x=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

8.

\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)-3sinx-cosx+2=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\) pt có đúng 1 nghiệm trên khoảng đã cho

\(2\cos^2x+\left(3-2m\right)\cos x=2-m\)

\(t=\cos x\Rightarrow t\in\left[-1;1\right]\)

\(\Delta=\left(3-2m\right)^2-4.2\left(m-2\right)=4m^2-12m+9-8m+16=4m^2-20m+25=\left(2m-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{2m-3-2m+5}{4}\\t=\frac{2m-3+2m-5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=m-2\end{matrix}\right.\)

\(t=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\left(tm\right)\)

Vậy để có 3 nghiệm thuộc khoảng (-pi/2;pi) thì pt còn lại cần 1 nghiệm nữa khác 2 nghiệm kia cũng thuộc khoảng (-pi/2;pi)

Xét hàm cos: \(t=m-2\) trong \(\left(-\frac{\pi}{2};\pi\right)\)

Nhìn vô bbt ta thấy \(-1< t< 0\) thì phương trình có 1 nghiệm

\(\Rightarrow-1< m-2< 0\Leftrightarrow1< m< 2\)

ĐKXĐ: ...

a/ \(\frac{sin2x}{cos2x}+\frac{cosx}{sinx}=8cos^2x\)

\(\Leftrightarrow sin2x.sinx+cos2x.cosx=8cos^2x.sinx.cos2x\)

\(\Leftrightarrow cosx=4sin2x.cos2x.cosx\)

\(\Leftrightarrow cosx=2sin4x.cosx\)

\(\Leftrightarrow cosx\left(2sin4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin4x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

b/ \(\frac{cosx}{sinx}-\frac{sinx}{cosx}+4sin2x=\frac{1}{sinx.cosx}\)

\(\Leftrightarrow cos^2x-sin^2x+4sin2x.sinx.cosx=1\)

\(\Leftrightarrow cos2x+2sin^22x=1\)

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

\(\Leftrightarrow-2cos^22x+cos2x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

1c/

\(5sinx-2=3\left(1-sinx\right)\frac{sin^2x}{1-sin^2x}\)

\(\Leftrightarrow5sinx-2=\frac{3sin^2x}{1+sinx}\)

\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)

\(\Leftrightarrow5sin^2x+3sinx-2=3sin^2x\)

\(\Leftrightarrow2sin^2x+3sinx-2=0\)

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

Bài 2:

a/ \(\Leftrightarrow\frac{\left(m+1\right)\left(1-cos2x\right)}{2}-sin2x+cos2x=0\)

\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(4+\left(m-1\right)^2\ge\left(m+1\right)^2\)

\(\Leftrightarrow4m\le4\Rightarrow m\le1\)

\(\Leftrightarrow2cos^2x-1-\left(2m-1\right)cosx-2m=0\)

\(\Leftrightarrow2cos^2x+cosx-1-2m\left(cosx+1\right)=0\)

\(\Leftrightarrow\left(cosx+1\right)\left(2cos-1\right)-2m\left(cosx+1\right)=0\)

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

\(\Leftrightarrow cosx=\frac{2m+1}{2}\)

Do \(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow0< cosx\le1\)

\(\Rightarrow0< \frac{2m+1}{2}\le1\Rightarrow-\frac{1}{2}< m\le\frac{1}{2}\)

Để pt làm sao bạn?

Câu 1:

\(cos7x-\sqrt{3}sin7x=-2\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow7x+\dfrac{\pi}{3}=-\pi+k2\pi\\ \Leftrightarrow x=-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\)

Vì \(x\in[\dfrac{2\pi}{5};\dfrac{6\pi}{7}]\)

\(\Rightarrow\dfrac{2\pi}{5}\le x\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{2\pi}{5}\le-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{31}{15}\le k\le\dfrac{11}{3}\)

Vì \(k\in Z\) nên \(k=3\)

Vậy \(x\) cần tìm là \(\dfrac{2\pi}{3}\)

Câu 2:

\(2sin^2x-sinxcosx-cos^2x=m\\ \Leftrightarrow2\dfrac{1-cos2x}{2}-\dfrac{1}{2}s\text{in2}x-\dfrac{1+cos2x}{2}=m\\ \Leftrightarrow3cos2x+s\text{in2}x=1-2m\)

Điều kiện để phương trình có nghiệm là:

\(3^2+1^2\ge\left(1-2m\right)^2\\ \Leftrightarrow4m^2-4m-9\le0\\ \Leftrightarrow\dfrac{1-\sqrt{10}}{2}\le m\le\dfrac{1+\sqrt{10}}{2}\)

thế 0 với pi vào pt ra 1 số âm 1 số dương => pht có 1 n.................................................................!

(0;pi)=(1;-8.89604401)

a: Khi m=3/2 thì phương trình sẽ trở thành:

\(cos^2x-\left(2\cdot\frac32+1\right)\cdot cosx+\frac32+1=0\)

=>\(cos^2x-4\cdot cosx+2,5=0\)

=>\(cos^2x-4\cdot cosx+4-1,5=0\)

=>\(\left(cosx-2\right)^2=1,5\)

=>\(\left[\begin{array}{l}cosx-2=\sqrt{1,5}\\ cosx-2=-\sqrt{1,5}\end{array}\right.\Rightarrow\left[\begin{array}{l}cosx=2+\sqrt{1,5}\\ cosx=2-\sqrt{1,5}\end{array}\right.\)

=>x∈∅