Chủ đề này có 3 trả lời

[ductuMATHER]

1, $\left ( 1+cosx \right )\left ( cos2x+2cosx \right ) + 2sin^{2}x = 0$

 

2, $sin^{2}4x-cos^{2}6x=sin(10x+\frac{21\Pi }{2})$

 

3, $sin^{2}(\frac{3x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{5x}{2}) = sin^{2}(\frac{11x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{13x}{2})$

[sharker]

1, $\left ( 1+cosx \right )\left ( cos2x+2cosx \right ) + 2sin^{2}x = 0$

 

2, $sin^{2}4x-cos^{2}6x=sin(10x+\frac{21\Pi }{2})$

 

3, $sin^{2}(\frac{3x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{5x}{2}) = sin^{2}(\frac{11x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{13x}{2})$

Bài 1

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

 $\Leftrightarrow (1 + \cos x)(2{\cos ^2}x - 1 + 2\cos x) + 2(1 - {\cos ^2}x) = 0$

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

 $\Leftrightarrow (\cos x + 1)(\cos 2x - 2) = 0$

[sharker]

1, $\left ( 1+cosx \right )\left ( cos2x+2cosx \right ) + 2sin^{2}x = 0$

 

2, $sin^{2}4x-cos^{2}6x=sin(10x+\frac{21\Pi }{2})$

 

3, $sin^{2}(\frac{3x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{5x}{2}) = sin^{2}(\frac{11x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{13x}{2})$

Bài 2:

$2{\sin ^2}4x - 2{\cos ^2}6x = 2\sin (10x + \frac{\pi }{2} + 10\pi )$

 $\Leftrightarrow (1 - \cos 8x) - (1 + \cos 12x) = 2\cos 10x$

 $\Leftrightarrow \cos 8x + \cos 12x + 2\cos 10x = 0$

 $\Leftrightarrow 2\cos 10x.\cos 2x + 2\cos 10x = 0$

 $\Leftrightarrow 2\cos 10x(\cos 2x + 1) = 0$

[sharker]

1, $\left ( 1+cosx \right )\left ( cos2x+2cosx \right ) + 2sin^{2}x = 0$

 

2, $sin^{2}4x-cos^{2}6x=sin(10x+\frac{21\Pi }{2})$

 

3, $sin^{2}(\frac{3x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{5x}{2}) = sin^{2}(\frac{11x}{2}) + sin^{2}(\frac{\Pi }{4}-\frac{13x}{2})$

Bài 3:

$2{\sin ^2}\frac{{3x}}{2} + 2{\sin ^2}\left( {\frac{\pi }{4} - \frac{{5x}}{2}} \right) = 2{\sin ^2}\frac{{11x}}{2} + 2{\sin ^2}\left( {\frac{\pi }{4} - \frac{{13x}}{2}} \right)$

 $\Leftrightarrow 1 - \cos 3x + 1 - \cos \left( {\frac{\pi }{2} - 5x} \right) = 1 - \cos 11x + 1 - \cos \left( {\frac{\pi }{2} - 13x} \right)$

 $\Leftrightarrow \cos 3x + \cos 5x = \cos 11x + \cos 13x$

 $\Leftrightarrow 2\cos 4x.\cos x = 2\cos 12x.\cos x$

 $\Leftrightarrow \cos x(\cos 4x - \cos 12x) = 0$

 $\Leftrightarrow 2\cos x.\sin 8x.\sin 4x = 0$