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2[sin6A+cos6A]-3[sin4A+cos4A]+1=0

Answer» Sorry, put these in LHS
Sin^6A+cos^6A=(sin^2A+cos^2A)^3-3sin^2Acos^2A(sin^2A+cos^2A)=1-3sin^2Acos^2And. Sin^4A+cos^4A=(sin^2A+cos^2A)^2-2sin^2Acos^A=1-2sin^2Acos^2APut these values in Rhs
2[sinA +sin (90-A)] - 3[ sin4A + sin(90-4A)] +12[1] -3[1] +1=0


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