Question
Tue November 13, 2012 By:

{(?3)sinx + cosx }

no. of solution of the equation are?

Expert Reply
Tue November 20, 2012
We know that asin?+bcos?=?a^2+b^2 sin(?+A)    where cos A=a/?a^2+b^2
 
Therefore, we have:

(?3)sinx + cosx) =2 sin(x+?/6)

And (?3)sin2x - cos2x) = 2 sin (2x-?/6)

And so

{(?3)sinx + cosx }t=4

where t = ? {(?3)sin2x - cos2x + 2}

Can be written as 2^t sin(x+?/6)^t=2^2

Where t=?2sin(2x-?/6)+2

 
This equation will have solution only when both sin(x+?/6) andsin(2x-?/6) will be one so that t=2and as this is possible for x=?/3,and sinx and sin2x both will be having common period of 2?leanth,solution will repeat after a length of 2? .so there will be  infinite solutions.
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