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Question

The solution of the equation 8sinx=3cosx+1sinx is/are given by

  1. x=nπ+π6, nZ.

  2. x=nππ6, nZ.

  3. x=nπ2+π12, nZ.

  4. x=nπ2π12, nZ.


Solution

The correct options are
A
x=nπ+π6, nZ.
D
x=nπ2π12, nZ.
Here, the given equation:
8sinx=3cosx+1sinx
Now, we can write the equation as:
8sinx=3sinx+cosxsinx cosx8sin2xcosx=3sinx+cosx4(2sinxcosx)sinx=                            3sinx+cosx4sin2xsinx=3sinx+cosx2(2sin2xsinx)=                            3sinx+cosx2(cos(2xx)cos(2x+x))                            =3sinx+cosx2cosx2cos3x                             =3sinx+cosxcosx2cos3x=3sinxcosx3sinx=2cos3xNow, dividing both sides by 2,we get12cosx32sinx=cos3xcosπ3cosxsinπ3sinx                                 =cos3xcos(x+π3)=cos3xThus, the general solution of theabove equation can be given as:3x=2nπ±(x+π3), nZ.Now, taking positive sign, we get3x=2nπ+(x+π3), nZ.2x=2nπ+π3, nZ.x=nπ+π6, nZ.Also, taking negative sign, we get:3x=2nπ(x+π3), nZ.4x=2nππ3, nZ.x=nπ2π12, nZ.
 

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