Question

How do you find the least solution to the equation `tan(3x-1)+2=5` in the interval [-5, -2]. the angle x is measured in radians?

Answer 1

The answer are: `x~=-2.4, -3.45, -4.5`

`tan(3x-1)+2=5rArrtan(3x-1)=3rArr`

`3x-1=arctan3+kpirArr3x=arctan3+1+kpirArr`

`x=(arctan3+1)/3+kpi/3=0,75+k1,05`.

So we need to give to `k` negative values since we want the solutions in `[-5,-2]`.

`k=-1rArrx=0.75-1.05~=-0.3` isn't in `[-5,-2]`,

`k=-2rArrx=0.75-2*1.05~=-1.35` isn't in `[-5,-2]`,

`k=-3rArrx=0.75-3*1.05~=-2.4` is in `[-5,-2]`,

`k=-4rArrx=0.75-4*1.05~=-3.45` is in `[-5,-2]`,

`k=-5rArrx=0.75-5*1.05~=-4.5` is in `[-5,-2]`,

`k=-6rArrx=0.75-6*1.05~=-5.55` isn't in `[-5,-2]`.

As you can control seeing the interceptions from the graph of the function `y=tan(3x-1)-3` and the x-axis.

graph{tan(3x-1)-3 [-6, 1, -5, 5]}