If &fnof;(x)=[cosx cos(x+2)-cos 2 (x+1)],where[.] denotes the greatest integer function≤x.Then solution of the equation &fnof;(x)=x is

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If &fnof;(x)=[cosx cos(x+2)-cos 2 (x+1)],where[.] denotes the greatest integer function&le;x.Then solution of the equation &fnof;(x)=x is

anandpal shekhawat · Answer

hi, [cos(x) cos(x+2) - cos^2(x+1)] = x expending above eqn & solving we get [(cos^2x)(cos2)-(cos^2x)cos^2(1) - sin^2(1) + cos^2x sin^2(1)] = x [(cos^2x){cos2 - cos^2(1) + sin^2(1)} - sin^2(1)] = x [(cos^2x){(2cos^2(1)-1) - cos^2(1) + (1-cos^2(1)} -sin^2(1)] = x [cos^x * 0 - sin^2(1)] = -1 = x becoz [-sin^2(1)] = -1 so, ans. is x = -1

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If ƒ(x)=[cosx cos(x+2)-cos 2 (x+1)],where[.] denotes the greatest integer function≤x.Then solution of the equation ƒ(x)=x is

If ƒ(x)=[cosx cos(x+2)-cos²(x+1)],where[.] denotes the greatest integer function≤x.Then solution of the equation ƒ(x)=x is

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