Pt lượng giác 11

X

xlovemathx

Đề thế này hả ?
a) [TEX] 2cos^{2} \frac{3x}{5} +1 = 3cos.\frac{4x}{5}[/TEX]
b) [TEX]1 + sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x = 2cos^{2}(\frac{\pi}{4} - \frac{x}{2}) [/TEX]

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T

thesunshine_after_rain

Đề thế này hả ?
a) [TEX] 2cos^{2} \frac{3x}{5} +1 = 3cos.\frac{4x}{5}[/TEX]
b) [TEX]1 + sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x = 2cos^{2}(\frac{\pi}{4} - \frac{x}{2}) [/TEX]

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[TEX]1 + sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x = 2cos^{2}(\frac{\pi}{4} - \frac{x}{2}) [/TEX]

[TEX]1- 2cos^{2}(\frac{\pi}{4} - \frac{x}{2})+ sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x =0[/TEX]

[TEX]cos(\frac{\pi}{2}-x)+sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x =0[/TEX]

[TEX]sinx+sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x =0[/TEX]

[TEX]sinx(1+sin.\frac{x}{2} - cos\frac{x}{2}sinx) =0[/TEX]
 
N

newtons007

Đề thế này hả ?
a) [TEX] 2cos^{2} \frac{3x}{5} +1 = 3cos.\frac{4x}{5}[/TEX]
b) [TEX]1 + sin.\frac{x}{2} sin x - cos\frac{x}{2}sin^{2}x = 2cos^{2}(\frac{\pi}{4} - \frac{x}{2}) [/TEX]

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a. [TEX] \Leftrightarrow 2 + cos(\frac{6x}{5})=3cos(\frac{4x}{5})[/TEX]
[TEX]\Leftrightarrow 4cos^3(\frac{2x}{5})-3cos^2(\frac{2x}{5})-3cos(\frac{2x}{5})+5=0[/TEX]
[TEX]\Leftrightarrow.......[/TEX]
 
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