Question #28e21

#8sinx=4+cosx#

#=>#Dividing by #8# in both sides.

#=>(8sinx)/8=(4+cosx)/8#

#=>(cancelcolor(red)(8)sinx)/cancelcolor(red)(8)=(4+cosx)/8#

#=>sinx=(4+cosx)/8 or# On simplifying

#=>sinx=4/8+cosx/8#

#=>sinx=(1*4)/(2*4)+cosx/8#

#=>sinx=(1*cancelcolor(red)(4))/(2*cancelcolor(red)(4))+cosx/8#

#=>sinx=1/2+cosx/8#

So answer = #sinx=(4+cosx)/8 or sinx= 1/2+cosx/8#

#8sinx=4+cosx#

#=>(8sinx-4)^2=cos^2x#

#=>64sin^2x-64sinx+16=1-sin^2x#

#=>65sin^2x-64sinx+15=0#

#=>65sin^2x-39sinx-25sinx+15=0#

#=>13sinx(5sinx-3)-5(5sinx-3)=0#

#=>(5sinx-3)(13sinx-5)=0#

So #sinx =3/5 and sin x= 5/13#

When #sinx =3/5# then #cosx =pmsqrt(1-sin^2x)=pm4/5#

#sinx=3/5 and cosx=4/5# when #x
in " 1st quadrant "
If we put these two values in the given equation we get

#LHS=8sinx=8xx3/4=24/5#
and

#RHS=4+cosx=4+4/5=24/5#

Here #LHS=RHS#

So we can say that #sinx =3/5# satisfies the given equation and this value is an acceptable solution when angle x is in 1st quadrant,

If x is in 2nd quadrant the #sinx =3/5 # but #cosx= -4/5#, these values do not satisfy the given equation

Again

When #sinx =5/13# then #cosx =pmsqrt(1-sin^2x)=pm12/13#

#sinx =5/13 and cosx =12/13# when #x in " 1st quadrant"#

If we put these two values in the given equation we get

#LHS=8sinx=8xx5/13=40/13#
and

#RHS=4+cosx=4+12/13=64/13#

Here #LHS!=RHS#

But if x is in 2nd quadrant the #sinx =5/13 # but #cosx= -12/13#, these values do satisfy the given equation

Hence solution #color(red)(tosinx =3/5" when x is in 1st qudrant") #

and #color(red)(tosinx =5/13" when x is in 2nd quadrant") #

Given that, #8sinx=4+cosx.#

#:. 8sinx-4=cosx.#

#:. (8sinx-4)^2=cos^2x=1-sin^2x.#

#:. 64sin^2x-64sinx+16+sin^2x-1=0, i.e., #

# 65sin^2x-64sinx+15=0.#

Applying the Quadr. Formula, we get,

#sinx={64+-sqrt(64^2-4*65*15)}/(2*65),#

#=(64+-sqrt(4096-3900))/130=(64+-sqrt196)/130=(64+-14)/130.#

#rArr sinx=78/130, or, 50/130.#

# :. sinx=3/5, or, 5/13.#