
Mark M. answered 02/05/21
Mathematics Teacher - NCLB Highly Qualified
tan 6x = tan 3x
tan (2(3x)) = tan 3x
(2 tan 3x) / (1 - tan2 3x) = tan 3x
2 / (1 - tan2 3x) = 1
2 = 1 - tan2 3x
1 = -tan2 3x
-1 = tan2 3x
∅
Abhi P.
asked 02/05/21Find the solutions of the equation that are in the interval [0, π/2). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
tan 6x = tan 3x
x=
Mark M. answered 02/05/21
Mathematics Teacher - NCLB Highly Qualified
tan 6x = tan 3x
tan (2(3x)) = tan 3x
(2 tan 3x) / (1 - tan2 3x) = tan 3x
2 / (1 - tan2 3x) = 1
2 = 1 - tan2 3x
1 = -tan2 3x
-1 = tan2 3x
∅
Let z = 3x to get tan2z = tanz
2tanz / (1 - tan2z) = tanz
2tanz = tanz - tan3z
tan3z + tanz = 0
tanz (tan2z + 1) = 0
tanz = 0
z = 0 , π , 2π , ...
3x = 0 , π , 2π , ...
x = 0 , π / 3 , ... (those are the only 2 solutions in the given interval)
Patrick B. answered 02/05/21
Math and computer tutor/teacher
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