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

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Answer 1 · 2015-03-04T20:12:15

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

$tan (3 x - 1) + 2 = 5 \Rightarrow tan (3 x - 1) = 3 \Rightarrow$ $tan(3x-1)+2=5rArrtan(3x-1)=3rArr$

$3 x - 1 = arctan 3 + k π \Rightarrow 3 x = arctan 3 + 1 + k π \Rightarrow$ $3x-1=arctan3+kpirArr3x=arctan3+1+kpirArr$

$x = \frac{arctan 3 + 1}{3} + k \frac{π}{3} = 0, 75 + k 1, 05$ $x=(arctan3+1)/3+kpi/3=0,75+k1,05$ .

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

$k = - 1 \Rightarrow x = 0.75 - 1.05 ≅ - 0.3$ $k=-1rArrx=0.75-1.05~=-0.3$ isn't in $[- 5, - 2]$ $[-5,-2]$ ,

$k = - 2 \Rightarrow x = 0.75 - 2 \cdot 1.05 ≅ - 1.35$ $k=-2rArrx=0.75-2*1.05~=-1.35$ isn't in $[- 5, - 2]$ $[-5,-2]$ ,

$k = - 3 \Rightarrow x = 0.75 - 3 \cdot 1.05 ≅ - 2.4$ $k=-3rArrx=0.75-3*1.05~=-2.4$ is in $[- 5, - 2]$ $[-5,-2]$ ,

$k = - 4 \Rightarrow x = 0.75 - 4 \cdot 1.05 ≅ - 3.45$ $k=-4rArrx=0.75-4*1.05~=-3.45$ is in $[- 5, - 2]$ $[-5,-2]$ ,

$k = - 5 \Rightarrow x = 0.75 - 5 \cdot 1.05 ≅ - 4.5$ $k=-5rArrx=0.75-5*1.05~=-4.5$ is in $[- 5, - 2]$ $[-5,-2]$ ,

$k = - 6 \Rightarrow x = 0.75 - 6 \cdot 1.05 ≅ - 5.55$ $k=-6rArrx=0.75-6*1.05~=-5.55$ isn't in $[- 5, - 2]$ $[-5,-2]$ .

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

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

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