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Section 1.6 : Solving Trig Equations with Calculators, Part II

9. Find all the solutions to \(10{x^2}\sin \left( {3x + 2} \right) = 7x\sin \left( {3x + 2} \right)\). Use at least 4 decimal places in your work.

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Hint : Factor the equation and using basic algebraic properties get some equations that can be dealt with using known techniques.
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Notice that if we move all the terms to one side we can then factor an \(x\) and a sine out of the equation. Doing this gives,

\[\begin{align*}10{x^2}\sin \left( {3x + 2} \right) - 7x\sin \left( {3x + 2} \right) & = 0\\ x\left( {10x - 7} \right)\sin \left( {3x + 2} \right) & = 0\end{align*}\]

Now, we have a product of three factors that equals zero and so by basic algebraic properties we know that we must have,

\[x = 0, \hspace{0.25in} 10x - 7 = 0,\hspace{0.25in}{\rm{OR}}\hspace{0.25in}\sin \left( {3x + 2} \right) = 0\]

Be careful with this type of equation to not make the mistake of just canceling the \(x\) or the sine from both sides in the initial step. Had you done that you would have missed the \(x = 0\) solution and the solutions we will get from solving \(\sin \left( {3x + 2} \right) = 0\).

When solving equations it is important to remember that you can’t cancel anything from both sides unless you know for a fact that what you are canceling will never be zero.

Hint : Solve each of these three equations to attain all the solutions to the original equation.
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There really isn’t anything that we need to do with the first equation and so we can move right on to the second equation (which also doesn’t really present any problems). Solving the second equation gives,

\[x = \frac{7}{{10}}\]

Now let’s take a look at the third equation. This equation is similar to equations solved earlier in this section. Therefore, we will be assuming that you can recall the solution process for each and we will not be putting in as many details. If you are unsure of the process you should go back to the previous section and work some of the problems there before proceeding with the solution to this problem.

From a unit circle we can see that we must have,

\[3x + 2 = 0 + 2\pi n\hspace{0.25in}{\rm{OR}}\hspace{0.25in}3x + 2 = \pi + 2\pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]

Notice that we can further reduce this down to,

\[3x + 2 = \pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]

Finally, the solutions from this equation are,

\[x = \frac{{\pi n - 2}}{3}\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]

Putting all of this together gives the following set of solutions.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0, \hspace{0.25in} x = \frac{7}{{10}}, \hspace{0.25in} {\rm{OR}} \hspace{0.25in} x = \frac{{\pi n - 2}}{3}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}\]

If an interval had been given we would next proceed with plugging in values of \(n\) to determine which solutions fall in that interval. Since we were not given an interval this is as far as we can go.

Note that depending upon the amount of decimals you used here your answers may vary slightly from these due to round off error. Any differences should be slight and only appear around the 4th decimal place or so however.