How do you find the general solutions for #tan^2x-tanx=0#?

1 Answer
Apr 2, 2018

The solutions are #x=pik, qquadpi/4+pik#

Explanation:

You can use a substitution, then solve the problem like a regular polynomial.

#tan^2x-tanx=0#

#(tanx)^2-tanx=0#

Now, let #u=tanx#:

#u^2-u=0#

#u(u-1)=0#

#u=0,1#

Plug #tanx# back in for #u#:

#tanx=0,tanx=1#

We know that #tanx# is #sinx/cosx#, so we can use that and solve each equation independently:

#color(white){color(black)( (sinx/cosx=0,qquadsinx/cosx=1), (sinx=0,qquadsinx=cosx), (x=0", "pi", "2pi", "3pi..., qquad???):}#

To figure out when #sinx# equals #cosx#, we can look at a unit circle:

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We can see that #sinx=cosx# when the angle is #pi/4,(5pi)/4,(9pi)/4...#

To write the general expression for #0,pi,2pi,3pi...# we can use the letter #k# (or #n#, depending on the person) to represent "any integer".

This pattern would be #x=pik#, because #sinx=0# when #x# is any multiple of #pi#.

The other pattern (#pi/4#, #(5pi)/4#, #(9pi)/4#...) can be rewritten as #pi/4+0pi,pi/4+1pi,pi/4+2pi...#. Now we can write a general expression for this using #k#: #x=pi/4+pik#

This means that the final solutions are:

#x=pik, qquadpi/4+pik#

Hope this helped!