sin^5x+cos^8x=1

Question

How to solve it?

Answer 1

sin x = a, cos x=b a^5+b^8=1 >a^5+[b^2]^4=1 >a^5+[1-a^2]^4=1 >a^5+a^8-4a^6+6a^4-4a^2+1=1 >a^8-4a^6+a^5+6a^4-4a^2=0 >a^2[a-1][a^5+a^4-3a^3-2a^2+4a+4]=0 >a=1,0 As a can't be 0 because if a=0 then perpendicular=0 which is not possible so a=1 where perpendicular=1 and hypotenuse=1 and b=0 where base =0 and hypotenuse=1 Therefore, sin x=1 cos x=0 By Safal Das Biswas Class X Student of S.A.V.M Chandernagore Live in Chinsurah