Giải phương trình lượng giác

[patranopcop]

[Toán 11] pt lượng giác

1,[tex]3cosx(1-cos2x)+2sin2x+sinx+cos2x=0 [/tex]

2,[tex]sin^3x+\sqrt{3}cos^3x+cos2x=sinxcos^2x+\sqrt{3}sin^2xcosx [/tex]

3,[tex]\frac{(1+2cosx)sinx}{(1-2cosx)(1+cosx)}=\sqrt{3} [/tex]

4,[tex]\frac{2cosx-2cos^3x+\sqrt{3}sin3x}{cos2x}=cosx-2 [/tex]

5,[tex]\frac{\sqrt{3}cos3x-4sinxcos^2x}{cosx}=\sqrt{3} [/tex]

giúp mình nha , tks nhiều !!

 

[trantien.hocmai]

$\text{câu 5} \\
PT \leftrightarrow \sqrt{3}(4\cos ^3x-3\cos x)-4\sin x.\cos ^2x=\sqrt{3} \cos x$

 

[trantien.hocmai]

$\text{câu 3} \\
PT \leftrightarrow \sin x+ \sin 2x=\sqrt{3}(1-\cos x-2\cos ^2x) \\
\leftrightarrow \sin x+\sqrt{3}\cos x=-\sqrt{3}\cos 2x-\sin 2x$

 

[trantien.hocmai]

$\text{câu 2} \\
\sin ^3x+\sqrt{3}\cos ^3x+\cos 2x-\sin x.\cos ^2x-\sqrt{3}\sin ^2x.\cos x=0 \\
\leftrightarrow -\sin x.\cos 2x+\sqrt{3}\cos x.\cos 2x+\cos 2x=0$

 

[buivanbao123]

cÂU 3)
\Leftrightarrow $Sinx+Sin2x=$\sqrt{3}(1-cosx-2cos^{2}x)$
\Leftrightarrow $Sinx+\sqrt{3}cosx+sin2x+cos2x=0$
Đến đây là pt cơ bản aSinx+bCosx=c

 

[trantien.hocmai]

$\text{câu 4} \\
PT \leftrightarrow \sin 2x.\sin x+\sqrt{3}\sin 3x=\cos 2x.\cos x-2\cos 2x \\
\leftrightarrow -\cos 3x+\cos x+2\sqrt{3}\sin 3x=\cos 3x+\cos x-4\cos 2x \\
\leftrightarrow 2\sqrt{3}\sin 3x-2\cos 3x=-4\cos 2x$

 

[patranopcop]

[Toán 11] pt lượng giác

1,[tex]\frac{8(sin^6x+cos^6x)(sinx-cosx)}{4-3sin^22x}+sin2x-1=0 [/tex]

2,[tex]\frac{sin^6x+cos^6x+sin^4x+cos^4x-2cos2x}{5cos2x-3}=0 [/tex]

3,[tex]1+(1+sin^2x)cosx=sin2x+sinx(1+cos^2x) [/tex]

4,[tex] sin^23x.cos2x+sin^2x=0 [/tex]

5,[tex] sinx.tanx+sin2x=tanx [/tex]

giúp mình nha , tks nhiều !

 

[patranopcop]

[Toán 11] pt lượng giác

1,[tex](1-sin^2x)cosx-(1-cos^2x)sinx=sin2x-1 [/tex]

2,[tex](sin{x\over2}-cos{x\over2})^2+\sqrt{3}cosx=1+\sqrt{2} [/tex]

3,[tex]\frac{1}{cos(x-\frac{\pi}{2})}-\frac{1}{sin(\frac{3\pi}{2}-x)}=4cos(x-\frac{5\pi}{4}) [/tex]

4,[tex] 2cos^2x+cosx=1-cos7x [/tex]

giúp mình nha , tks nhiều !

 

[trantien.hocmai]

$\text{câu 2} \\
PT \leftrightarrow 1-\frac{3}{4}\sin ^22x+1-\frac{1}{2} \sin ^22x-2\cos 2x=0 \\
\leftrightarrow 2-\frac{5}{4}(1-\cos ^22x)-2\cos 2x=0$

 

[trantien.hocmai]

$\text{câu 1} \\
\leftrightarrow 8(1-\frac{3}{4}\sin ^22x)(\sin x-\cos x)-(1-\sin 2x)(4-3\sin ^22x)=0 \\
\leftrightarrow 8(1-\frac{3}{4}\sin ^22x)(\sin x-\cos x)-(\sin x-\cos x)^2(3-3\sin ^22x)=0$

 

[trantien.hocmai]

$\text{câu 5} \\
PT \leftrightarrow \sin ^2x+2\sin x.\cos ^2x-\sin x=0$

 

[xuanquynh97]

Bài 1 ĐK

PT \Leftrightarrow $\dfrac{8(1-3sin^2xcos^2x)(sinx-cosx}{4-3sin^22x}+sin2x-1=0$

\Leftrightarrow $\dfrac{8(1-3sin^2xcos^2x)(sinx-cosx)}{4(1-3sin^2xcos^2x)}+sin2x-1=0$

\Leftrightarrow $2(sinx-cosx)+sin2x-1=0$

 

[trantien.hocmai]

$\text{câu 3} \\
PT \leftrightarrow (\sin x-\cos x)^2+\cos x+\sin ^2x.\cos x-\sin x-\sin x-\sin x.\cos ^2x=0 \\ \leftrightarrow (\sin x-\cos x)^2-(\sin x-\cos x)+\sin x.\cos x(\sin x-\cos x)=0$

 

[xuanquynh97]

Bài 3 : $1+(1+sin^2x)cosx=sin2x+sinx(1+cos^2x) $

PT \Leftrightarrow $cosx-sinx+sinxcosx(sinx-cosx)+1-2sinxcosx=0$

Đặt $sinx-cosx=t$

\Rightarrow $-t+\dfrac{1-t^2}{2}.t+1-1+t^2=0$

 

[trantien.hocmai]

$\text{câu 2} \\
PT \leftrightarrow 1-\sin x+\sqrt{3}\cos x=1+\sqrt{2} \\
\leftrightarrow \sqrt{3}\cos x-\sin x=\sqrt{2}$

 

[xuanquynh97]

Bài 4 :

$\dfrac{1-cos6x}{2}.cos2x+\dfrac{1-cos2x}{2}=0$

\Leftrightarrow $1-cos2xcos6x=0$

\Leftrightarrow $\dfrac{1}{2}(cos8x+cos4x)−1=0$

\Leftrightarrow $2cos^24x+cos4x−3=0$

 

[trantien.hocmai]

$\text{câu 3} \\
\text{đặt }t=\frac{3\pi}{2}-x \rightarrow x=\frac{3\pi}{2}-t \\
\text{thay vào cac góc kia rồi làm nhá}$

 

[mua_sao_bang_98]

1. $(1-sin^2x)cosx-(1-cos^2x)sinx=sin2x-1$

\Leftrightarrow $cos^3x-sin^3x=sin2x-1$

\Leftrightarrow $(cosx-sinx)(1+sinxcosx)+(cosx-sinx)^2=0$

\Leftrightarrow $(cosx-sinx)(1+sinxcosx+cosx-sinx)=0$

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

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

(2):Đặt sinxcosx = a , cosx-sinx=b \Rightarrow $b^2=1-2a$ (3)

pt \Rightarrow $1+a+b=0$ (4)

Từ (3) (4) \Rightarrow $\left[\begin{matrix} cosx-sinx=-1 (5)\\ cosx-sinx=3 (Vô nghiệm) \end{matrix}\right.$

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

 

[quangminhh]

LƯỢNG GIÁC hay

$\sqrt{\sin x}+ \sin x + \dfrac{1-\cos 2x}{2}+\cos x = 1$

mọi người giúp đỡ

Chú ý gõ latex: http://diendan.hocmai.vn/showthread.php?t=247845

 

[levietdung1998]

đặt điều kiện cho sin x >0
$\begin{array}{l}
\begin{array}{*{20}{c}}
{\sqrt {\sin x} + \sin x + \frac{{1 - \cos 2x}}{2} + \cos x = 1} \\
{ < = > \sqrt {\sin x} + \sin x + \frac{{1 - (2{{\cos }^2}x - 1)}}{2} + \cos x = 1} \\
{ < = > \sqrt {\sin x} + \sin x + 1 - c{\rm{o}}{{\rm{s}}^2}x + \cos x = 1} \\
{ < = > \sqrt {\sin x} + \sin x - c{\rm{o}}{{\rm{s}}^2}x + \cos x = 0} \\
{ < = > \left( {\sqrt {\sin x} + \cos x} \right).\left( {1 + \sqrt {\sin x} - \cos x} \right) = 0} \\
{} \\
\end{array} \\
< = > \left[ \begin{array}{l}
\sqrt {\sin x} = - \cos x \\
\sqrt {\sin x} = \cos x + 1 \\
\end{array} \right. \\
\end{array}$