Question

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

Answer 1

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`

Answer 2

`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.!