First slide
Trigonometric equations
Question

The A.M of all the solutions of 4cos3x4cos2xcos(π+x)+1=0in(0,315)is

Difficult
Solution

4cos3x4cos2xcos(x+π)1=0 

4cos3x4cos2x+cosx1=0 

4cos2xcosx-1+1cosx1=0 

cosx=1   hassolutionsasx=2π,4π,8π,....100π  in  (0,315) 

 

A.M=sumofnumbersno.ofnumbers=2π+4π+......+100π50=51π

 

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