[ Toán 11] Phương trình lượng giác

N

ng.mai_96

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N

nguyenbahiep1

câu 1

[laTEX]sin ( 2x + 2\pi + \frac{\pi}{2}) - 3cos( x - 3.\pi - \frac{\pi}{2}) = 1 + 2sin x \\ \\ cos2x + 3.sinx = 1 + 2sin x \\ \\ cos2x -1 + sin x = 0 \\ \\ 1 - 2sin^2 x -1 +sin x = 0 \\ \\ sin x( 2sin x - 1) = 0 [/laTEX]
 
H

huutho2408

chào bạn

b, $\dfrac{sin^2 2x + 4cos^4 2x - 1}{\sqrt{2sinxcosx}}=0$
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ĐK: $sin2x\not=0$$\Longleftrightarrow x\not=\dfrac{k\pi}{2}$

$\dfrac{sin^2 2x + 4cos^4 2x - 1}{\sqrt{2sinxcosx}}=0$

$\Longleftrightarrow sin^2 2x + 4cos^4 2x - 1=0$

$\Longleftrightarrow 1-cos^2 2x + 4cos^4 2x - 1=0$

$\Longleftrightarrow 4cos^4 2x-cos^2 2x =0$

$\Longleftrightarrow cos^2 2x.(4cos^2 2x -1) =0$
 
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