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?

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Answer 1 · 2015-03-04T20:12:15

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]}