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

2 Answers
Cesareo R.
Apr 29, 2017

See below.

Explanation:

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

Ratnaker Mehta
Apr 29, 2017

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

Explanation:

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.!

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