Question

How do you solve the following equation `tan(3t)=sqrt(3)` in the interval [0, 2pi]?

Answer 1

Solution is `t={pi/9,(4pi)/9,(7pi)/9,(10pi)/9,(13pi)/9,(16pi)/9}`

Explanation:

We have `tan(pi/3)=sqrt3` and as the ratio tangent repeats after every `pi`,

Solution of `tan(3t)=sqrt3` has solution

`3t={pi/3,(4pi)/3,(7pi)/3,(10pi)/3,(13pi)/3,(16pi)/3,(19pi)/3,(22pi)/3......}` and

`t={pi/9,(4pi)/9,(7pi)/9,(10pi)/9,(13pi)/9,(16pi)/9,(19pi)/9,(22pi)/9......}`.

But as we seek solution in interval `0<=t<=2pi`

Solution is `t={pi/9,(4pi)/9,(7pi)/9,(10pi)/9,(13pi)/9,(16pi)/9}`