How to solve this equation? #cosx+sqrt(3)sinx=a^2;x inRR#,and for what values of #a#,the equation has solutions?

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Answer 1 · 2017-04-29T16:07:27

See below.

#alpha cos x+beta sinx = a^2#

or

#alpha/sqrt(alpha^2+beta^2)cosx+beta/sqrt(alpha^2+beta^2)sinx = a^2/sqrt(alpha^2+beta^2)#

calling now

#alpha/sqrt(alpha^2+beta^2)=cos phi#
#beta/sqrt(alpha^2+beta^2)=sin phi#

#cosphi cosx+sin phi sinx = cos(x-phi) = a^2/sqrt(alpha^2+beta^2)#

Here #alpha = 1, beta = sqrt3# so

#phi = pi/3#

then

#cos(x-pi/3)=a^2/2#

but #-1 le cos x le 1# so feasible values for #a# must obey

#-1 le a^2/2 le 1# or

#0 le a^2 le 2-> abs a le sqrt2#

with those conditions

#x-pi/3=arccos(a^2/2)+2kpi# with #k in ZZ#

or

#x = arccos(a^2/2)+pi/3+2kpi# with #k in ZZ#

Answer 2 · 2017-04-29T16:19:29

#x=2kpi+pi/6+-arc cos(a^2/2), k in ZZ iff |a| lesqrt2.#

#1cosx+sqrt3sinx=a^2.#

Dividing the eqn. by, #sqrt{(1)^2+(sqrt3)^2}=2,#we get,

#1/2cosx+sqrt3/2sinx=a^2/2, i.e.,#

#cos(pi/3)cosx+sin(pi/6)sinx=a^2/2, or, #

#cos(x-pi/6)=a^2/2#

# because, |costheta| le 1, AA theta in RR," the Soln. exists "iff a^2/2 le 1.#

Hence, the Soln. Set exists, # iff |a| le sqrt2.#

Under this cond.,

#cos(x-pi/6)=a^2/2 rArr x-pi/6=2kpi+-arc cos(a^2/2), k in ZZ, or, #

#x=2kpi+pi/6+-arc cos(a^2/2), k in ZZ.#

Enjoy Maths.!