Question

How do you find all solutions of the equation `cos(x+pi/6)-cos(x-pi/6)=1` in the interval `[0,2pi)`?

Answer 1

See below.

Explanation:

`cos(x + a) - cos(x - a) = -2 sin x sin a`

so

`-2 sin x sin a=1->sinx = -1/(2sina) = -1/(2 xx 1/2) = -1`

so

`x = arcsin(-1)+2kpi=- pi/2+2kpi` then

`x = 3/2pi`

Answer 2

`x = (3pi)/2`

Explanation:

Here's a different way of proceeding. Use the sum and difference formulae `cos(A + B) = cosAcosB - sinAsinB` and `cos(A - B) = cosAcosB + sinAsinB`.

`cosxcos(pi/6) - sinxsin(pi/6) - (cosxcos(pi/6) + sinxsin(pi/6)) = 1`

`cosx(sqrt(3)/2) - sinx(1/2) - (cosx(sqrt(3)/2) + sinx(1/2)) = 1`

`sqrt(3)/2cosx - 1/2sinx - sqrt(3)/2cosx - 1/2sinx = 1`

`-sinx = 1`

`sinx = -1`

`x = (3pi)/2`

Hopefully this helps!