pt lượng giác

N

nguyenbahiep1

[laTEX]t = tan(\frac{x}{2}) \\ \\ \frac{2t}{1+t^2} + 2\frac{2t}{1+t^2}.\frac{1-t^2}{1+t^2} - 1 = 0 \\ \\ 2t(1+t^2) + 4t(1-t^2) - (1+t^2)^2 = 0 \\ \\ (1-t)(t^3+3t^2+5t-1) = 0 \\ \\ t = 1 \Rightarrow x = \frac{\pi}{2} + k2\pi[/laTEX]
 
You must log in or register to reply here.
Top Bottom