sin 2x = 1而砌师訶君2323sin3x + cos3x +可 nHHTrf而言

1.	sin 2x = 1而砌师訶君2323sin3x + cos3x +可 nHHTrf而言
Answer» 1 + sin^3x + cos^3x =3/2*sin2x 1+(sinx+cosx)(sin^2x-sinxcosx+cos^2x)=3sinxcosx 1+(sinx+cox)(1-sinxcosx)=3sinxcosx Let s=sinx+cosx s²=1+2sinxcosx 1+s(1-(1-s²/2))=3(1-s²/2) s^3+3s^2-3s-5=0 (s+1)(s^2+2s-5)=0 range of s is [-√2,√2] s=-1 s=√2(sin(x+π/4))=-1 (sin(x+π/4))=-1/√2 x+π/4=-π/4+n(-1)^nπ

Answer»

1 + sin^3x + cos^3x =3/2*sin2x

1+(sinx+cosx)(sin^2x-sinxcosx+cos^2x)=3sinxcosx

1+(sinx+cox)(1-sinxcosx)=3sinxcosx

Let s=sinx+cosx

s²=1+2sinxcosx

1+s(1-(1-s²/2))=3(1-s²/2)

s^3+3s^2-3s-5=0

(s+1)(s^2+2s-5)=0

range of s is [-√2,√2]

s=-1

s=√2(sin(x+π/4))=-1

(sin(x+π/4))=-1/√2

x+π/4=-π/4+n(-1)^nπ

Discussion