Uni2805
Làm cho em bài toán?
4sin^2(pi + x/2) - căn3.sin(9pi/2 - 2x) = 1 + 2cos^2(x - 3pi/4)
truongson
20/05/2013 23:26
sinxcos4x - sin^22x = 4sin^2(frac{pi}{4} - frac{x}{2}) - frac{7}{2}
Ủng hộ nào, thax cái nha
sinxcos4x - sin^22x = 4sin^2(frac{pi}{4} - frac{x}{2}) - frac{7}{2}
Rightarrow sinx.cosx+frac{cos4x}{2}-frac{1}{2}+2sinx+frac{3}{2}=2-2.cos(frac{pi}{2}-x)-frac{7}{2}
Rightarrow sinx.cosx-frac{1}{2}.(1-cos4x)+2sinx+frac{3}{2}=0
Rightarrow sinxcos4x+2sinx+frac{1}{2}cos4x+1=0
Rightarrow (sinx+frac{1}{2}).(cos4x+2)=0
Rightarrow left[
thang_star
20/05/2013 23:26
sinxcos4x - sin^22x = 4sin^2(frac{pi}{4} - frac{x}{2}) - frac{7}{2}
Ủng hộ nào, thax cái nha
sinxcos4x - sin^22x = 4sin^2(frac{pi}{4} - frac{x}{2}) - frac{7}{2}
Rightarrow sinx.cosx+frac{cos4x}{2}-frac{1}{2}+2sinx+frac{3}{2}=2-2.cos(frac{pi}{2}-x)-frac{7}{2}
Rightarrow sinx.cosx-frac{1}{2}.(1-cos4x)+2sinx+frac{3}{2}=0
Rightarrow sinxcos4x+2sinx+frac{1}{2}cos4x+1=0
Rightarrow (sinx+frac{1}{2}).(cos4x+2)=0
Rightarrow left[
