### Video Transcript

In this video, we will learn how to
solve a trigonometric equation using factoring or squaring. The equations that weβll look at
will involve at least one of the trigonometric functions sine, cosine, and
tangent. Before looking at these new
methods, we will refamiliarize ourselves with solving simple trigonometric
equations.

We recall that equations of the
form sin π equals π, cos of two π equals π, tan of π minus 30 equals π can all
be solved using either a graph or a CAST diagram. We should also recall some key
properties relating to the sine, cosine, and tangent functions. The sin of angle π is equal to the
sin of 180 degrees minus π. The cos of π degrees is equal to
the cos of 360 degrees minus π. And the tan of any angle π is
equal to the tan of 180 degrees plus π.

Before solving a trigonometric
equation, it is often helpful to consider how many solutions we expect the equation
to have. We can identify the number of times
a trigonometric function is equal to a particular value in a given interval by
drawing a horizontal line across its graph at this value and then counting the
number of times this line intersects the graph. For example, if we want to
determine the number of solutions to the equation sin π₯ equals 0.5 in the interval
where π₯ is greater than or equal to zero and less than or equal to 360 degrees, we
draw a horizontal line at π¦ equals 0.5. As this horizontal line crosses the
graph twice in the given interval, we can conclude that the equation sin π₯ equals
0.5 has two solutions in the interval π₯ is greater than or equal to zero and less
than or equal to 360 degrees.

We will now look at a more complex
example of this type of problem. Before doing so, we also need to
recall one of the key trigonometric identities. This identity defines the
relationship between the three trig functions. For any angle π, the tan of π is
equal to the sin of π divided by the cos of π. We will now look at an example
where we need to use this identity.

If π₯ is greater than or equal to
zero degrees and less than or equal to 360 degrees, then the number of solutions of
the equation four sin π₯ is equal to the tan of π₯ is what.

This equation involves two trig
ratios: sine and tangent. We recall that we can express the
tangent function in terms of the sine and cosine functions. The tan of π₯ is equal to the sin
of π₯ over the cos of π₯. Substituting this into the
right-hand side of our equation, we have four sin π₯ is equal to sin π₯ over cos
π₯. Next, we can subtract sin π₯ over
cos π₯ from both sides, giving us four sin π₯ minus sin π₯ over cos π₯ equals
zero.

At this stage, we might be tempted
to divide the equation through by the shared factor of sin π₯. However, doing this could
potentially result in the loss of some solutions if the factor we divide by is equal
to zero. This means that instead we will
factor sin π₯ out of the left-hand side of the equation. This gives us sin π₯ multiplied by
four minus one over cos π₯ is equal to zero.

We now have a product that is equal
to zero. And the only way a product can
equal zero is if at least one of the factors itself equals zero. This means that we need to solve
the two equations sin π₯ equals zero and four minus one over cos π₯ equals zero. Recalling the graph of the sine
function as shown, we see that sin π₯ is equal to zero three times in the interval
where π₯ is greater than or equal to zero and less than or equal to 360 degrees. These solutions are zero, 180, and
360 degrees. However, in this question, we are
only interested in the number of solutions. sin π₯ is equal to zero three times
between zero and 360 degrees inclusive.

Letβs now consider the second
equation, four minus one over cos π₯ equals zero. Multiplying through by cos of π₯
gives us four cos of π₯ minus one equals zero. We can then add one to both sides
such that four cos π₯ equals one and, finally, divide through by four such that the
cos of π₯ is equal to one-quarter.

Recalling the graph of the cosine
function and drawing a horizontal line across the graph at π¦ equals one-quarter, we
find that there are two values of π₯ in the interval π₯ is greater than or equal to
zero and less than or equal to 360 degrees for which the cos of π₯ equals
one-quarter. Whilst we could calculate these
exact values, we donβt need to in this question. However, it is clear that these are
not the same values for which sin π₯ equals zero, as one of the solutions lies
between zero and 90 degrees and the second solution lies between 270 and 360
degrees.

We can therefore conclude that
there are five solutions to the equation four sin π₯ equals tan π₯ between zero and
360 degrees inclusive. The correct answer is five.

We will now look at an example
where we need to find all the solutions of a more complex trigonometric equation by
factoring.

Find the set of values satisfying
tan squared π plus tan π equals zero, where π is greater than or equal to zero
degrees and less than 180 degrees.

Upon inspection, we notice that
this is a quadratic equation in tan π. We can begin by factoring the
left-hand side of our equation. This gives us tan π multiplied by
tan π plus one is equal to zero. Setting each of these factors equal
to zero, we have tan π equals zero or tan π plus one equals zero. Subtracting one from both sides of
our second equation, we now have two solutions: the tan of π equals zero or the tan
of π equals negative one.

Next, we recall the graph of the
tangent function as shown. Whilst weβve drawn it for values of
π between zero and 360 degrees, it is important to note in this question that we
are only looking for solutions greater than or equal to zero and less than 180
degrees. We can see from the graph that the
tan of π equals zero at zero degrees. This is also true for 180 and 360
degrees. However, these two values are not
in the set of values for π. The equation tan π equals zero
therefore has one solution where π equals zero degrees.

If we draw a horizontal line on our
graph at π¦ equals negative one, we see that this intersects the graph of π¦ equals
tan π once between zero and 180 degrees. This value lies somewhere between
90 and 180 degrees. Considering the equation tan π
equals negative one, we can take the inverse tangent of both sides. This gives us π is equal to the
inverse tan of negative one.

Typing this into our calculator
gives us π is equal to negative 45 degrees. This value is, however, outside the
required interval for π. Recalling the periodicity of the
tangent function, we can find the second solution to the equation by adding 180
degrees. This is because the tan of π is
equal to the tan of 180 degrees plus π. We need to add 180 to negative
45. This gives us an answer of 135
degrees, which does lie within the required interval for π. This is therefore a second valid
solution.

From the graph, we can see that
there are no further solutions such that π is greater than or equal to zero and
less than 180 degrees. We can therefore conclude that the
set of values that satisfy tan squared π plus tan π equals zero are zero degrees
and 135 degrees.

It is important to note at this
stage that a common mistake would be to divide the initial equation by tan π. This would mean, however, we would
lose one of the solutions to the original equation, as it is possible for tan π to
equal zero. Indeed, this was one of the
equations we were subsequently required to solve. We should also take careful note of
the interval in which weβre looking for solutions. It is common for the interval to be
π is greater than or equal to zero and less than or equal to 360 degrees. However, as in this question, this
is not always the case. Additional values may well be valid
solutions to the equation. But if they are out of the
specified interval, then they are not correct in the context of the problem.

Before looking at one final
example, we need to consider a second trigonometric identity. This identity is known as the
Pythagorean identity. And it states that for all values
of π, sin squared π plus cos squared π equals one. This identity can be used to solve
some specific types of trigonometric equations. In the final example, we will need
to use this after squaring both sides of the equation. It is important to note, however,
that squaring both sides of an equation can be risky if we do not take great
care. This is because squaring and
finding the square root are not one-to-one operations. When squaring, we can create an
extra solution. Therefore, if we need to solve a
trigonometric equation by squaring, we must subsequently check all our solutions in
the original equation to ensure we have not obtained any extraneous values.

By first squaring both sides, or
otherwise, solve the equation four sin π minus four cos π is equal to the square
root of three, where π is greater than zero degrees and less than or equal to 360
degrees. Be careful to remove any extraneous
solutions. Give your answers to two decimal
places.

The question advises that we
approach the problem by first squaring both sides of the equation. Doing so, we obtain four sin π
minus four cos π all squared is equal to root three squared. Distributing the parentheses and
then collecting like terms on the left-hand side gives us 16 sin squared π minus 32
sin π cos π plus 16 cos squared π. On the right-hand side, root three
squared is equal to three.

Next, we recall the Pythagorean
identity, which states that sin squared π plus cos squared π is equal to one. Simplifying the left-hand side
further gives us 16 multiplied by sin squared π plus cos squared π minus 32 sin π
cos π. Replacing sin squared π plus cos
squared π with one gives us the equation 16 minus 32 sin π cos π equals
three. Subtracting 16 from both sides of
this equation gives us negative 32 sin π cos π is equal to negative 13. We can then divide through by
negative 32 such that sin π cos π equals 13 over 32.

We now have two equations in the
two variables sin π and cos π. This means that the system of
equations can be solved simultaneously. Adding four cos π to both sides of
our original equation, we have four sin π is equal to root three plus four cos
π. Dividing both sides of this
equation by four, we have sin π is equal to root three plus four cos π all divided
by four.

After clearing some space, we will
now consider how we can solve these two simultaneous equations. We will begin by substituting the
expression for sin π in equation two into equation one. This gives us root three plus four
cos π over four multiplied by cos π is equal to 13 over 32. We can simplify this equation by
firstly distributing the parentheses. We can then multiply through by 32,
giving us eight root three cos π plus 32 cos squared π equals 13. Finally, subtracting 13 from both
sides of this equation, we have the quadratic equation in terms of cos π as
shown.

This can be solved using the
quadratic formula, where π is 32, π is eight root three, and π is negative
13. Substituting in these values and
then simplifying gives us cos of π is equal to negative three plus or minus the
square root of 29 all divided by eight. Taking the inverse cosine of both
sides with positive root 29 gives us π is equal to 62.829 and so on. To two decimal places, this is
equal to 62.83 degrees. Taking the inverse cosine of our
equation with negative root 29 gives us π is equal to 152.829 and so on. This rounds to 152.83 degrees to
two decimal places.

We were asked to give all solutions
that are greater than or equal to zero degrees and less than or equal to 360
degrees. We therefore need to consider the
symmetry of the cosine function such that the cos of π is equal to the cos of 360
degrees minus π. Subtracting each of our values from
360 degrees gives us further solutions 297.17 degrees and 207.17 degrees to two
decimal places.

We have therefore found four
possible solutions to the given equation. However, we were reminded in the
question to remove any extraneous solutions, these extra solutions that were created
when we squared our original equation. We need to substitute each of our
four solutions into the initial equation to check they are valid.

The initial equation was four sin
π minus four cos π equals root three. Substituting π equals 62.83 into
the left side of our equation gives us an answer of root three. This means that this is a valid
solution. However, when we substitute π is
equal to 152.83 degrees into the left-hand side of our equation, we do not get root
three. This means that this is not a valid
solution. Repeating this process for 207.17
degrees and 297.17 degrees, we see that 207.17 is a valid solution, whereas the
fourth answer of 297.17 is not. We can therefore conclude that
there are two solutions that satisfy the equation in the given interval of π, which
are 62.83 and 207.17 degrees.

We will now summarize the key
points from this video. We saw in this video that some
trigonometric equations can be solved by factoring. It is extremely important that any
common factors are factored by and not divided by, as this will avoid the potential
loss of solutions if these factors are equal to zero. We also saw that some trigonometric
equations can be solved by squaring both sides. Whenever we use this approach, it
is important to avoid creating any extraneous solutions. We can check for these by
substituting in any solutions back in to the original equation.

We saw that the two identities tan
of π is equal to the sin of π divided by the cos of π and sin squared π plus cos
squared π equals one can be useful in solving trigonometric equations. Finally, the graphs of
trigonometric functions, their properties, and the CAST diagram can be used to
determine additional solutions following the determination of a principal angle.