Question

How do you find the general solutions for `sqrt{3}(sin(x)) + cos(x) = 1`?

Answer 1

`x = (6n+(-1)^n-1)pi/6, " " n in ZZ`

Explanation:

`R = sqrt(a^2+b^2)`
`a sin x + b cos x = R(a/R sin x + b/R cos x) = R sin (x + alpha)`
where `a/R = cos alpha` and `b/R = sin alpha` such that `alpha = arctan (b/a)`

`sqrt(3) sin x + cos x = 2 sin (x + pi/6) = 1`
`sin (x + pi/6) = 1/2`

`x + pi/6 = npi + (-1)^npi/6, " " n in ZZ`
`x = (6n+(-1)^n-1)pi/6, " " n in ZZ`