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1 + sin^3x + cos^3x =3/2 sin2x. Find general solution of x.

Khushi goyal , 7 Years ago
Grade 11
anser 1 Answers
Akash Bhattacharya

Last Activity: 7 Years ago

this answer is very simple
1+sin^3x+cos^3x=3/2sin2x
1+(sinx+cosx)(sin^2x+cos^2x-sinxcosx)=3sinxcosx(a^3-b^3= (a+b)(a62+b^2-ab) & sin2x=2sinxcosx)
let sinx+cosx=t
s.o.b.s (sinx+cosx)^2=t^2
sin^2x+cos^2x+2sinxcosx=t^2
1+2sinxcosx=t^2 (sin^2x+cos^2x=1)
sinxcosx=t^2/2-1/2
so now the equation is after solving is t^3+3t^2-3t-5=0
after doing synthetic division you will get sin(x+pi/4)=1or 5^1/2-1/2^1/2
and get the solutions respectively.
 

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