Question

Sin cube x + cos cube x

Answer 1

Step-by-step explanation:

sin^3x+cos^3x=1-3sinx cosx

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

(sinx+cosx) (1-sinx cosx) = 1-3sinx cosx multiply X 2

(sinx+cosx) (2-2sinx cosx)= 2-6sinx cosx

(sinx+cosx) (2-sin2x) = 2 - 3sin2x now square both sides

(sin^2x + cos^2x + 2sinx cosx) (2-sin2x)^2 =(2- 3sin2x)^2

(1+sin2x)(4-4sin2x + sin^2[2x])=4 -12sin2x +9sin^2[2x]

4-4sin[2x]+ sin^2[2x]+ 4sin[2x]- 4sin^2[2x]+sin^3[2x] =4-12sin[2x] + 9sin^2[2x]

sin^3[2x] -12 sin^2[2x] +20 sin[2x]=0

sin2x(sin^2[2x]-12sin[2x] +20)=0

sin2x(sin2x - 10)(sin2x -2)=0

sin2x=0

sin2x=10 (not possible)>1)

or sin2x=2 (not possible>1)

so the only possible solution

is sin2x=0 --->2x= 0 or π

so x=0 or π/2

Answer 2

x=0 or pi/2 is the answer