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### Section 1.6 : Solving Trig Equations with Calculators, Part II

Because this document is also being prepared for viewing on the web we split this section into two parts to keep the size of the pages to a minimum.

Also, as with the last few examples in the previous part of this section we are not going to be looking for solutions in an interval in order to save space. The important part of these examples is to find the solutions to the equation. If we’d been given an interval it would be easy enough to find the solutions that actually fall in the interval.

In all the examples in the previous section all the arguments, the \(3t\), \(\frac{\alpha }{7}\), *etc*., were fairly simple. Let’s take a look at an example that has a slightly more complicated looking argument.

Example 1 Solve \(5\cos \left( {2x - 1} \right) = - 3\).

Show Solution

Note that the argument here is not really all that complicated but the addition of the “-1” often seems to confuse people so we need to a quick example with this kind of argument. The solution process is identical to all the problems we’ve done to this point so we won’t be putting in much explanation. Here is the solution.

\[\cos \left( {2x - 1} \right) = - \frac{3}{5}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}2x - 1 = {\cos ^{ - 1}}\left( { - \frac{3}{5}} \right) = 2.2143\]

This angle is in the second quadrant and so we can use either -2.2143 or \(2\pi - 2.2143 = 4.0689\) for the second angle. As usual for these notes we’ll use the positive one. Therefore the two angles are,

\[\begin{align*}\begin{aligned}2x - 1 & = 2.2143 + 2\pi n\\ 2x - 1 & = 4.0689 + 2\pi n\end{aligned} &{\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

Now, we still need to find the actual values of \(x\) that are the solutions. These are found in the same manner as all the problems above. We’ll first add 1 to both sides and then divide by 2. Doing this gives,

\[\begin{align*}\begin{aligned}x & = 1.6072 + \pi n\\ x & = 2.5345 + \pi n\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

So, in this example we saw an argument that was a little different from those seen previously, but not all that different when it comes to working the problems so don’t get too excited about it.

We now need to move into a different type of trig equation. All of the trig equations solved to this point (the previous example as well as the previous section) were, in some way, more or less the “standard” trig equation that is usually solved in a trig class. There are other types of equations involving trig functions however that we need to take a quick look at. The remaining examples show some of these different kinds of trig equations.

Example 2 Solve \(2\cos \left( {6y} \right) + 11\cos \left( {6y} \right)\sin \left( {3y} \right) = 0\).

Show Solution

So, this definitely doesn’t look like any of the equations we’ve solved to this point and initially the process is different as well. First, notice that there is a \(\cos \left( {6y} \right)\) in each term, so let’s factor that out and see what we have.

\[\cos \left( {6y} \right)\left( {2 + 11\sin \left( {3y} \right)} \right) = 0\]

We now have a product of two terms that is zero and so we know that we must have,

\[\cos \left( {6y} \right) = 0\hspace{0.5in}{\rm{OR}}\hspace{0.5in}2 + 11\sin \left( {3y} \right) = 0\]

Now, at this point we have two trig equations to solve and each is identical to the type of equation we were solving earlier. Because of this we won’t put in much detail about solving these two equations.

First, solving \(\cos \left( {6y} \right) = 0\) gives,

\[\begin{align*}\begin{aligned}6y & = \frac{\pi }{2} + 2\pi n\\ & \\ 6y & = \frac{{3\pi }}{2} + 2\pi n\end{aligned} & {\hspace{0.25in} \Rightarrow \hspace{0.25in}} &\begin{aligned}y & = \frac{\pi }{{12}} + \frac{{\pi n}}{3}\\ & \\ y & = \frac{\pi }{4} + \frac{{\pi n}}{3}\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

Next, solving \(2 + 11\sin \left( {3y} \right) = 0\) gives,

\[\begin{align*}\begin{aligned}3y & = 6.1004 + 2\pi n\\ & \\3y & = 3.3244 + 2\pi n\end{aligned} & {\hspace{0.25in} \Rightarrow \hspace{0.25in}} &\begin{aligned}y & = 2.0335 + \frac{{2\pi n}}{3}\\ & \\ y & = 1.1081 + \frac{{2\pi n}}{3}\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

Remember that in these notes we tend to take positive angles and so the first solution here is in fact \(2\pi - 0.1828\) where our calculator gave us -0.1828 as the answer when using the inverse sine function.

The solutions to this equation are then,

\[\begin{align*}\begin{aligned}y & = \frac{\pi }{{12}} + \frac{{\pi n}}{3}\\ & \\ y & = \frac{\pi }{4} + \frac{{\pi n}}{3}\\ y & = 2.0335 + \frac{{2\pi n}}{3}\\ y & = 1.1081 + \frac{{2\pi n}}{3}\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

This next example also involves “factoring” trig equations but in a slightly different manner than the previous example.

Example 3 Solve \(4{\sin ^2}\left( {\frac{t}{3}} \right) - 3\sin \left( {\frac{t}{3}} \right) = 1\).

Show Solution

Before solving this equation let’s solve an apparently unrelated equation.

\[4{x^2} - 3x = 1\hspace{0.5in} \Rightarrow \hspace{0.25in}4{x^2} - 3x - 1 = \left( {4x + 1} \right)\left( {x - 1} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - \frac{1}{4},1\]

This is an easy (or at least we hope it’s easy at this point) equation to solve. The obvious question then is, why did we do this? We’ll, if you compare the two equations you’ll see that the only real difference is that the one we just solved has an \(x\) everywhere and the equation we want to solve has a sine. What this tells us is that we can work the two equations in exactly the same way.

We, will first “factor” the equation as follows,

\[4{\sin ^2}\left( {\frac{t}{3}} \right) - 3\sin \left( {\frac{t}{3}} \right) - 1 = \left( {4\sin \left( {\frac{t}{3}} \right) + 1} \right)\left( {\sin \left( {\frac{t}{3}} \right) - 1} \right) = 0\]

Now, set each of the two factors equal to zero and solve for the sine,

\[\sin \left( {\frac{t}{3}} \right) = - \frac{1}{4}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\sin \left( {\frac{t}{3}} \right) = 1\]

We now have two trig equations that we can easily (hopefully…) solve at this point. We’ll leave the details to you to verify that the solutions to each of these and hence the solutions to the original equation are,

\[\begin{align*}\begin{aligned}t & = 18.0915 + 6\pi n\\ t & = 10.1829 + 6\pi n\\ t & = \frac{{3\pi }}{2} + 6\pi n\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{align*}\]

The first two solutions are from the first equation and the third solution is from the second equation.

Let’s work one more trig equation that involves solving a quadratic equation. However, this time, unlike the previous example this one won’t factor and so we’ll need to use the quadratic formula.

Example 4 Solve \(8{\cos ^2}\left( {1 - x} \right) + 13\cos \left( {1 - x} \right) - 5 = 0\).

Show Solution

Now, as mentioned prior to starting the example this quadratic does not factor. However, that doesn’t mean all is lost. We can solve the following equation with the quadratic formula (you do remember this and how to use it right?),

\[8{t^2} + 13t - 5 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = \frac{{ - 13 \pm \sqrt {329} }}{{16}} = 0.3211,\,\,\, - 1.9461\]

So, if we can use the quadratic formula on this then we can also use it on the equation we’re asked to solve. Doing this gives us,

\[\cos \left( {1 - x} \right) = 0.3211\hspace{0.25in}\hspace{0.25in}{\rm{OR}}\hspace{0.25in}\hspace{0.25in}\cos \left( {1 - x} \right) = - 1.9461\]

Now, recall Example 9 from the previous section. In that example we noted that \( - 1 \le \cos \left( \theta \right) \le 1\) and so the second equation will have no solutions. Therefore, the solutions to the first equation will yield the only solutions to our original equation. Solving this gives the following set of solutions,

\[\begin{array}{cc}\begin{aligned}x & = - 0.2439 - 2\pi n\\ x & = - 4.0393 - 2\pi n\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

Note that we did get some negative numbers here and that does seem to violate the general form that we’ve been using in most of these examples. However, in this case the “-” are coming about when we solved for \(x\) after computing the inverse cosine in our calculator.

There is one more example in this section that we need to work that illustrates another way in which factoring can arise in solving trig equations. This equation is also the only one where the variable appears both inside and outside of the trig equation. Not all equations in this form can be easily solved, however some can so we want to do a quick example of one.

Example 5 Solve \(5x\tan \left( {8x} \right) = 3x\).

Show Solution

First, before we even start solving we need to make one thing clear. **DO NOT CANCEL AN \(x\) FROM BOTH SIDES!!!** While this may seem like a natural thing to do it **WILL** cause us to lose a solution here.

So, to solve this equation we’ll first get all the terms on one side of the equation and then factor an \(x\) out of the equation. If we can cancel an \(x\) from all terms then it can be factored out. Doing this gives,

\[5x\tan \left( {8x} \right) - 3x = x\left( {5\tan \left( {8x} \right) - 3} \right) = 0\]

Upon factoring we can see that we must have either,

\[x = 0\hspace{0.5in}{\rm{OR}}\hspace{0.5in}\tan \left( {8x} \right) = \frac{3}{5}\]

Note that if we’d canceled the \(x\) we would have missed the first solution. Now, we solved an equation with a tangent in it in Example 5 of the previous section so we’ll not go into the details of this solution here. Here is the solution to the trig equation.

\[\begin{array}{cc}\begin{aligned}x & = 0.0676 + \frac{{\pi n}}{4}\\ & \\ x & = 0.4603 + \frac{{\pi n}}{4}\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

The complete set of solutions then to the original equation are,

\[\begin{array}{cc}\begin{aligned}x & = 0 \\ x & = 0.0676 + \frac{{\pi n}}{4}\\ & \\ x & = 0.4603 + \frac{{\pi n}}{4}\end{aligned} & {\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }\end{array}\]