Question

Question #d665b

Answer 1

`x=(3n+1)pi/9-2/3, n in ZZ`.

Explanation:

We will use :

The Soln. Set of the eqn. `: tantheta=tanalpha` is `theta in {npi+alpha, n in ZZ}`.

Since, `tan(3x+2)=sqrt3=tan(pi/3)`, we have,

`(3x+2)=npi+pi/3=(3n+1)pi/3 rArr 3x=(3n+1)pi/3-2`

`:. x=(3n+1)pi/9-2/3, n in ZZ`.