We're asked to solve the equations
over the interval from zero to two pi,
including zero radians and not including two pi radians.
The first equation is tangent x plus one equals zero.
We first isolate tangent x by subtracting one on both sides,
which gives us tangent x equals negative one.
Using the unit circle,
tangent theta is equal to y divided by x.
Since we want y divided by x to be equal to negative one,
we need to find the points on the unit circle
where the x and y-coordinates are opposites of one another.
When we divide opposites,
the quotient is always negative one.
So the first thing to recognize is that the terminal side
of the angles must be in quadrant two,
where x is negative and y is positive,
or in quadrant four, where x is positive and y is negative.
Looking in the second quadrant,
notice how, at this point here,
where x equals negative square root two divided by two
and y is equal to positive square root two divided by two,
y divided by x is going to equal negative one.
And therefore, the terminal side of the angle
must intersect the unit circle at this point here.
We know the initial side is along the positive x-axis.
And therefore, the angle from zero to two pi radians
would be this angle here, that measures 135 degrees
or three pi over four radians, or 3/4 pi radians.
So the first solution is x equals 3/4 pi radians,
and now we'll have another solution in the fourth quadrant.
Notice at this point on the unit circle,
where x is positive square root two divided by two
and y is negative square root two divided by two,
y divided by x is also negative one.
And therefore, the terminal side of the angle
must be this right here,
that intersects the unit circle at this point.
And therefore, in standard position,
the angle is going to be 315 degrees
or 7/4 pi radians, or seven pi over four radians.
So there are two solutions to the equation
over the given interval.
And now let's solve the second equation.
We are given cosecant x minus two equals zero.
The first step is to add two to both sides
to isolate cosecant x.
So we have cosecant x equals two.
On the unit circle, cosecant theta is equal to one over y,
which is the reciprocal of the sine function value.
And since we know that cosecant x equals two,
let's also find the sine function value.
We can write the cosecant function value of a fraction
with the denominator of one,
which means if cosecant x equals two over one,
then sine x must equal the reciprocal of two over one,
which is one over two, or one half.
It'll be easier to solve this equation here,
because, on the unit circle, sine theta is equal to y.
So to solve the given equation,
let's find where the y-coordinates on the unit circle
are equal to positive one half.
Well, we know y is positive in the first quadrant
and the second quadrant.
And notice the y-coordinate is positive one half
at this point of the unit circle.
And therefore, this ray would be the terminal side
of the angle in standard position.
We know the initial side is here.
And therefore, the first solution to the given equation
is x equals 30 degrees
or pi over six radians, or 1/6 pi radians.
So let's give the first solution as x equals 1/6 pi radians,
which again is commonly written as pi over six radians.
There will also be a second angle
with a terminal side in the second quadrant.
Notice at this point on the unit circle,
the y-coordinate is also positive one half.
And therefore, the sine function value is one half,
and the cosecant function value
would have to be two over one, or two.
And therefore, the terminal side of the angle
must be this right here, that intersects the unit circle
at this point where y equals one half.
In standard position, this angle is 150 degrees
or 5/6 pi radians, or five pi over six radians,
which means the second solution
over the given interval is x equals 5/6 pi radians.
I hope you found this helpful.