Question

# The solution of the equation 8sinx=√3cosx+1sinx is/are given by x=nπ+π6, n∈Z. x=nπ−π6, n∈Z. x=nπ2+π12, n∈Z. x=nπ2−π12, n∈Z.

Solution

## The correct options are A x=nπ+π6, n∈Z. D x=nπ2−π12, n∈Z.Here, the given equation: 8sinx=√3cosx+1sinx Now, we can write the equation as: 8sinx=√3sinx+cosxsinx cosx⇒8sin2xcosx=√3sinx+cosx⇒4(2sinxcosx)sinx=                            √3sinx+cosx⇒4sin2xsinx=√3sinx+cosx⇒2(2sin2xsinx)=                            √3sinx+cosx⇒2(cos(2x−x)−cos(2x+x))                            =√3sinx+cosx⇒2cosx−2cos3x                             =√3sinx+cosx⇒cosx−2cos3x=√3sinx⇒cosx−√3sinx=2cos3xNow, dividing both sides by 2,we get12cosx−√32sinx=cos3x⇒cosπ3cosx−sinπ3sinx                                 =cos3x⇒cos(x+π3)=cos3xThus, the general solution of theabove equation can be given as:3x=2nπ±(x+π3), n∈Z.Now, taking positive sign, we get3x=2nπ+(x+π3), n∈Z.⇒2x=2nπ+π3, n∈Z.⇒x=nπ+π6, n∈Z.Also, taking negative sign, we get:3x=2nπ−(x+π3), n∈Z.⇒4x=2nπ−π3, n∈Z.⇒x=nπ2−π12, n∈Z.

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