### Video Transcript

In this video, we will learn how to
find the general solution of a trigonometric equation or how to solve it over a
specified interval. A trigonometric equation is an
equation which involves at least one of the following: a trig function such as sine,
cosine, and tangent; a reciprocal trig function, cosecant, secant, and cotangent; or
an inverse of any of these. Some of the simpler examples of
such equations can be solved without using a calculator. In these cases, we will use our
knowledge of the special angles together with the symmetry and periodicity of the
sine, cosine, and tangent graphs.

We will begin this video by
recalling the exact values for the sine, cosine, and tangent of a number of special
angles. It is important that we are able to
recall the sine, cosine, and tangent of zero, 30, 45, 60, and 90 degrees. It is also important to know the
corresponding values of these angles in radians. Whilst we’ll not consider the
proofs of these in any detail in this video, it is important to recall that they
come from our knowledge of right-angle trigonometry and the Pythagorean theorem. The exact values of sine, cosine,
and tangent of the angles given are shown. Recalling the identity tan 𝜃 is
equal to sin 𝜃 over cos 𝜃, we can calculate the tangent of any of these angles by
dividing the value of the sine of the angle by the value of the cosine of the
angle.

In our first example, we will
demonstrate how to use the symmetry of the graph of the sine function alongside this
table of values to find all solutions to a simple trig equation.

What is the general solution of sin
𝜃 is equal to root two over two?

In order to find the general
solution to a trigonometric equation, we begin by finding a particular solution. In this case, the table of exact
trig values can help. For any angle 𝜃 given in radian
measure, the exact values of the sine function are as shown. We observe that the sin of 𝜋 by
four radians is equal to root two over two. This means that 𝜃 equals 𝜋 over
four is a particular solution to the equation sin 𝜃 equals root two over two. To find further solutions, we
sketch the graph of 𝑦 equals sin 𝜃 between zero and two 𝜋. The solutions to sin 𝜃 equals root
two over two are found by adding the line 𝑦 equals root two over two to the
diagram.

We notice that this intersects the
curve twice between zero and two 𝜋. The first point of intersection
corresponds to the solution 𝜋 over four. Since the sine curve has symmetry
about 𝜋 over two in the interval from zero to 𝜋, the second solution is found by
subtracting 𝜋 over four from 𝜋. This is equal to three 𝜋 over
four. We now have two solutions to the
equation sin 𝜃 equals root two over two: 𝜃 equals 𝜋 over four and 𝜃 equals three
𝜋 over four. Recalling that the sine function is
periodic with a period of 360 degrees or two 𝜋 radians, we can find the general
solution. Firstly, we have 𝜃 is equal to 𝜋
over four plus two 𝑛𝜋 — we can write it like this because further solutions are
found by adding or subtracting multiples of two 𝜋 or 360 degrees — and secondly,
three 𝜋 over four plus two 𝑛𝜋, where 𝑛 is an integer.

In this question, we demonstrated
how to interpret the symmetry of the graph of the sine function to find all
solutions to an equation. Another way of extending the domain
of the sine function is the unit circle. Recalling that the unit circle is
centered at the origin with a radius of one unit, we can work out the sine of any
angle 𝜃 by starting at the point one, zero and traveling along the circumference of
the circle in a counterclockwise direction until the angle that is formed between
this point, the origin, and the positive 𝑥-axis is equal to 𝜃. If this point has coordinates 𝑥,
𝑦, then sin 𝜃 is equal to the value of 𝑦. The value of the 𝑦-coordinate is
positive in both the first and second quadrants. Hence, the value of sin 𝜃 will
also be positive in these quadrants.

Since the unit circle has
reflection symmetry about the 𝑦-axis, we can see that sin 𝜃 is equal to sin of 180
degrees minus 𝜃. By continuing to move along the
circumference of the unit circle, we see that sin 𝜃 is also equal to sin of 360
plus 𝜃 for all values of 𝜃. These results can be generalized as
shown where the set of all solutions to sin 𝜃 equals 𝐶 is 𝜃 equals 𝜃 sub one
plus 360𝑛 and 𝜃 equals 180 minus 𝜃 sub one plus 360𝑛, where 𝑛 is an
integer. Note that if 𝜃 is measured in
radians, 360 degrees is replaced by two 𝜋 and 180 degrees by 𝜋. Whilst we might feel inclined to
memorize these formulae, in practice, it could be much more effective to sketch the
graph of the function or the unit circle.

In our next example, we will look
at how to use the symmetry of the graph of the cosine function to solve a
trigonometric equation.

Find the set of values satisfying
the cos of 𝜃 minus 105 is equal to negative a half, where 𝜃 is greater than zero
degrees and less than 360 degrees.

In order to find the solutions to a
trig equation in a given interval, we begin by finding a particular solution. In this case, the table of exact
trigonometric values can help. We will first redefine the argument
of the function by letting 𝛼 equal 𝜃 minus 105 such that the cos of 𝛼 is equal to
negative one-half and 𝜃 is equal to 𝛼 plus 105. We can then amend the interval over
which our solutions are valid by adding 105 to each part of the inequality; 𝛼 is
greater than 105 degrees and less than 465 degrees. Filling in the table for the exact
values of cos 𝛼, we can see that cos 𝛼 equals a half when 𝛼 is 60 degrees. However, there are no values of 𝛼
in the table such that cos 𝛼 is equal to negative a half.

By sketching the graph of the
cosine function along with the lines 𝑦 equals one-half and 𝑦 equals negative a
half, we can find the associated value of 𝛼. It appears on the graph that there
might be three values between 105 and 465 degrees. As the graph has rotational
symmetry between zero and 180 degrees about 90 degrees, zero, the first solution is
equal to 180 minus 60. This is equal to 120 degrees, which
lies in the required interval. Next, using symmetry of the curve,
we have 𝛼 is equal to 180 plus 60. This is equal to 240 degrees, which
also lies in the given interval. The third solution corresponds to
120 plus 360 degrees. However, this value of 480 degrees
lies outside of our interval for 𝛼. Hence, the solutions to cos 𝛼
equals negative one-half are 𝛼 equals 120 degrees and 𝛼 equals 240 degrees.

We can now calculate the
corresponding values of 𝜃. 120 plus 105 is equal to 225, and
240 plus 105 is 345. The set of values that satisfy cos
of 𝜃 minus 105 equals negative one-half are 225 degrees and 345 degrees. An alternative technique to find
the particular solution to cos of 𝛼 equals negative one-half is to use the inverse
cosine function such that 𝛼 is equal to the inverse cos of negative one-half, which
is equal to 120 degrees. From this point, we would use the
same steps to find the other solutions. This could also have been done
using the unit circle, which would lead us to the general rule: the cos of 𝜃 is
equal to the cos of 360 degrees minus 𝜃.

Using the symmetry of the unit
circle and periodicity of the cosine function, we can quote formulas for the general
solution to equations involving this function. In the same way as we have already
seen for the sine function, then the set of all solutions to cos 𝜃 equals 𝐶 is 𝜃
is equal to 𝜃 sub one plus 360𝑛 and 𝜃 is equal to 360 minus 𝜃 sub one plus 360𝑛
for all integer values of 𝑛. Once again, if 𝜃 is measured in
radians, we replace 360 degrees with two 𝜋 radians.

We saw in the previous question how
to solve a trig equation where the argument of the function has been transformed in
some way. We will now look at a similar
version of this involving the tangent function.

Find the set of values satisfying
the tan of two 𝑥 plus 𝜋 over five is equal to negative one, where 𝑥 is greater
than or equal to zero and less than or equal to two 𝜋.

To solve this equation, we’ll begin
by redefining the argument as this will allow us to use the symmetry of the tangent
function. We will let 𝜃 equal two 𝑥 plus 𝜋
over five. This means that we need to solve
the tan of 𝜃 equals negative one where 𝜃 is greater than or equal to 𝜋 over five
and less than or equal to 21𝜋 over five as we multiply each part of the inequality
by two and then add 𝜋 over five. Next, we recall that for 𝜃
measured in radians, the exact values of tan 𝜃 are as shown. We see that tan of 𝜋 over four is
equal to one. Next, we will sketch the graph of
𝑦 equals the tan of 𝜃. We will then add the horizontal
lines where 𝑦 equals one and 𝑦 equals negative one.

Due to the rotational symmetry of
the tangent function, the first solution occurs when 𝜃 is equal to 𝜋 minus 𝜋 over
four. This is equal to three 𝜋 over
four. As the function is periodic with a
period of 𝜋 radians, we can find the remaining solutions by adding multiples of 𝜋
to this value. Firstly, three 𝜋 over four plus 𝜋
is equal to seven 𝜋 over four. We also have solutions 11𝜋 over
four and 15𝜋 over four. These are the four points of
intersection shown on the graph. Clearing some space and rewriting
our four solutions for 𝜃, we can now calculate the values of 𝑥. As 𝜃 is equal to two 𝑥 plus 𝜋
over five, two 𝑥 is equal to 𝜃 minus 𝜋 over five. Dividing through by two, we have 𝑥
is equal to 𝜃 over two minus 𝜋 over 10.

We can now substitute each of our
values of 𝜃 into this equation. This gives us four values of 𝑥
equal to 11𝜋 over 40, 31𝜋 over 40, 51𝜋 over 40, and 71𝜋 over 40. This is the set of values that
satisfies the equation tan of two 𝑥 plus 𝜋 over five equals negative one where 𝑥
lies between zero and two 𝜋 inclusive.

As we have done for the sine and
cosine functions, we can now quote the general solutions to equations involving the
tangent function. When 𝜃 is measured in degrees, the
solutions are 𝜃 is equal to 𝜃 sub one plus 180𝑛 where 𝑛 is an integer. And if 𝜃 is measured in radians,
we have 𝜃 is equal to 𝜃 sub one plus 𝑛𝜋 where, once again, 𝑛 is an integer.

In this video, we’ve only
considered the standard trigonometric functions sine, cosine, and tangent. Whilst we’ll not cover them here,
it is important to understand that the process holds for the reciprocal functions
cosecant, secant, and cotangent.

We will now recap the key points
from this video. We can solve simple trigonometric
equations using tables of exact values or the inverse trig functions. To help us calculate all solutions
to a given equation in a specified range, we can draw the graph of the necessary
trig function or use the unit circle. The symmetry and periodicity of the
sine, cosine, and tangent functions allow us to calculate further solutions to
trigonometric equations or general solutions involving integer multiples of 360
degrees or two 𝜋 radians for sine and cosine and 180 degrees or 𝜋 radians for
tangent.