a.

\(\Leftrightarrow\left(1-sin^2x\right)\left(1+sin^2x\right)-\frac{5}{3}cos^4x=0\)

\(\Leftrightarrow cos^2x\left(1+sin^2x\right)-\frac{5}{3}cos^4x=0\)

\(\Leftrightarrow cos^2x\left(3+3sin^2x-5cos^2x\right)=0\)

\(\Leftrightarrow cos^2x\left(3+\frac{3}{2}-\frac{3}{2}cos2x-\frac{5}{2}-\frac{5}{2}cos2x\right)=0\)

\(\Leftrightarrow cos^2x\left(2-4cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

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

a)bung hằng đẳng thức số 3 ra còn 5/3cos^4(x) giữ lại

Sau đó (1-sin^2(x)) là cos^2x sau đó rút nhân tử chung là cos^2(x) ra ta được

cos^2(x)(1+sin^2(x)-5/3cos^2(x))=0

Cho từng vế = 0 rr giải

b)rút sin x ra nhưng giữ thg cos2x lại rr rút nhân tử chung là cos2x ta đc

cos2x(1-sinx)=0

Cho từng vế =0 rr giải

c)chém 4cos^2(x) ở hai vế hai bên thì chỉ còn

cos3x+6cosx=0 <=> 4cos^3(x)+3cosx=0

Bấm máy tìm cosx

1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)

\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)

\(\Leftrightarrow5\sin2x+5\cos2x=5\)

\(\Leftrightarrow\cos2x+\sin2x=1\)

\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)

2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)

\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)

\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)

\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)

\(\Leftrightarrow2cosx-sinx-4sin^2x.cosx+2sin^3x=sin^3x+cos^3x\)

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

- Với \(\left\{{}\begin{matrix}cosx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\) là nghiệm của pt

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

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

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

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{3}{2}\end{matrix}\right.\)

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

a)

\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)

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

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)

c)

\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)

\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)

d)

\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)

\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)

f)

\(\cos 2x-\cos 4x=0\)

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

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

b,e,g bạn xem lại đề, đơn vị không thống nhất.

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/

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

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

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

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

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

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

1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)

\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)

\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)

\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)

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

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

\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)

\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)

\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)

\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)

3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)

\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)

\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)

\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)

-TH1: ...

- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...

4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)

\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)

\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)

\(\Leftrightarrow\cos x+\sin x=-1\)

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)