[toán 10] lượng giác

H

hothithuyduong

giải phương trình:[TEX] 1) sin 2x= cos (x+ \frac{\pi}{6})[/TEX]

[TEX]\leftrightarrow cos(\frac{\pi}{2} - 2x) = cos(x+ \frac{\pi}{6})[/TEX]

[TEX]\leftrightarrow \left[\begin{\frac{\pi}{2} - 2x = x + \frac{\pi}{6} + k2\pi}\\{\frac{\pi}{2} - 2x = - x - \frac{\pi}{6} + k2\pi}[/TEX]

[TEX]\leftrightarrow \left[\begin{x = \frac{\pi}{9} - \frac{k2\pi}{3}}\\{x = \frac{2\pi}{3} + k2\pi}[/TEX]

[TEX]2) sin x= cos (2x+ \frac{\pi}{3})[/TEX]


Tương tự câu trên

[TEX]\leftrightarrow cos( \frac{\pi}{2} - x) = cos (2x+ \frac{\pi}{3}) [/TEX]




[TEX]\leftrightarrow 2sin^2x = 1[/TEX]


[TEX]\leftrightarrow 1 - 2sin^2x = 0 [/TEX]


[TEX]\leftrightarrow cos2x = 0[/TEX]


 
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