Here is the solution to the tangent problem we started in class: `3tan(1/2x)+3=0, x in [-pi/2, 4pi]`
Step 1: `3tan(1/2x)=-3`
Step 2: `tan(1/2x)=-1`
Here we need to take the inverse (actually arc) tangent.
Step 3: `1/2x=tan^-1(-1)`
Here you are thinking “What angle gives me a tangent of -1?” In other words, “when is sine divided by cosine equal to -1?” Your unit circle gives you the answer.
Step 4: `1/2x=(3pi)/4` or `1/2x=(7pi)/4`
Solving for x gives you `x=(3pi)/2` or `x=(7pi)/2`.
Lastly, adding or subtracting complete revolutions (periods) of this function will produce co-terminal angles which also have a tangent of -1. We want all possibilities between `-pi/2` and `4pi`.
Keep in mind that the parent tangent function has a period of `pi` and our equation has a horizontal dilation by a factor of `1//2`. This means we are fitting `1//2` of a cycle in `pi` radians or in other words a full cycle in `2pi` radians.
Adding `2pi` to `(3pi)/2` gives us `(7pi)/2` (of course!) and adding another `2pi` would be `(11pi)/2` which is bigger than our solution range of `4pi`. However, subtracting `2pi` from `(3pi)/2` gives us `-pi/2` which is in our solution range.
Therefore our final solution set is `{-pi/2, (3pi)/2, (7pi)/2}`.
Let’s check the graph for verification. (`1.6 = pi/2` on the graph)
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xmin=-pi/2; xmax=4pi; xscl=pi/2; axes();
plot(3tan(1/2x)+3);
endagraph
Our solutions are, in fact, where we said they should be.