We have tan θ on both sides of the equation. This is unacceptable, just like it wouldn't be in an equation with x. We'll start with fixing that through some basic algebra. Subtract (2 tan θ) from both sides, and then rearrange the 1 as well:
tan θ = 1
Much nicer. Now we are ready for the inverse:
θ = tan-1 (1)
We know that 45-45-90 triangles have two equal legs, so .
Our domain is restricted, so we don't have to go looking for every solution under the sun. Just the ones between 0 and 2π, which spend their days indoors reading Shmoop instead of playing outside. We like, but we also dislike; don't be a complete shut in on our account.
Tangent is positive in Quadrants I and III, so we have another angle that would work here in Quadrant III. It has a reference angle of , and great references from its last job, working at the front counter at the thunderdome.
We have solutions of and for the equation.
Example 2
Solve .
At first glance this may look like pure craziness, but don't go running just yet. How does this look?
2x2 + x = 1
Not nearly as bad, right? It is perfectly fine to start out by using x instead of writing out sine (or whatever trig function is there). It's easier to look at and will save us some headache. We can solve this one through factoring. Doing that nets us:
(2x – 1)(x + 1) = 0
When can break this up into two separate equations to solve:
2x – 1 = 0
2x = 1
x = ½
and
x + 1 = 0
x = -1
Now is the fated return of sine, and the fabled "finding of the inverses":
sin (θ) = ½
and
sin (θ) = -1
Taking the inverse of these (and remembering that sine is positive in Quadrants I and II and negative in Quadrants III and IV):
(Quadrant I), (Quadrant II)
and
Our solutions are only defined between 0 and 2π, so we have to convert our result there into one of its coterminal angles.
Putting all our solutions together gets us:
That problem was nuttier than a squirrel in a Jif factory, but that's what you get when you break the solution up into two separate equations. If we can handle this much, and we've shown we can, then we should be pretty comfortable with this section.
Example 3
Solve .
Don't worry too much about the 2θ in there to start. We'll get to it. The first step is to isolate the trig function, but, lucky us, that's already done here. Let's start with the inverse then:
Leave 2θ alone. Instead, find the angle for arcsine. The positive angle for ½ is . We want negative angles, though, and that means consulting ASTC (A Soft, Tender Cookie). It says we have two negative angles, in Quadrants III and IV.
We still can't divide by 2θ. There are an infinite number of solutions, and we want to divide them all by 2. That's pretty ambitious, but we can do it by setting up the equations:
, for n = 0, ±1, ±2, etc.
, for n = 0, ±1, ±2, etc.
And now the long-awaited division, and then the problem is done.
.
.
, and great references from its last job, working at the front counter at the thunderdome.
and 
for the equation.
.
(Quadrant I), 
(Quadrant II)
.
. We want negative angles, though, and that means consulting ASTC (A Soft, Tender Cookie). It says we have two negative angles, in Quadrants III and IV.
, for n = 0, ±1, ±2, etc.
, for n = 0, ±1, ±2, etc.
, for n = 0, ±1, ±2, etc.
, for n = 0, ±1, ±2, etc.