Which values of $a$ satisfy a trigonometric equation in a given interval?

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In summary, a trigonometric equation is an equation that contains one or more trigonometric functions and can be solved for values of the variable that make it true. An interval in a trigonometric equation is a range of values for the variable, and it can be determined using algebraic techniques and properties of trigonometric functions. It is possible for a trigonometric equation to have multiple solutions within a given interval due to the periodic behavior of trigonometric functions. Solutions can be checked by plugging them back into the original equation or using a graphing calculator or software.
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Find all $a$ in the interval $\left(0,\,\dfrac{\pi}{2}\right)$ such that $\dfrac{\sqrt{3}-1}{\sin a}+\dfrac{\sqrt{3}+1}{\cos a}=4\sqrt{2}$.


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Congratulations to the following members for their correct solutions::)

1. MarkFL
2.
greg1313
3. kaliprasad

Solution from MarkFL:
Because both $\sin(a)$ and $\cos(a)$ are non-zero on the given interval, we may multiply through by $\sin(a)\cos(a)$ to obtain:

\(\displaystyle \left(\sqrt{3}-1\right)\cos(a)+\left(\sqrt{3}+1\right)\sin(a)=4\sqrt{2}\sin(a)\cos(a)\)

Using a linear combination on the left and the double-angle identity for sine on the right, there results:

\(\displaystyle 2\sqrt{2}\sin\left(a+\frac{\pi}{12}\right)=2\sqrt{2}\sin(2a)\)

or:

\(\displaystyle \sin\left(a+\frac{\pi}{12}\right)=\sin(2a)\)

In light of the identity $\sin(\pi-x)=\sin(x)$, we have 2 cases to consider:

(i) \(\displaystyle \sin\left(a+\frac{\pi}{12}\right)=\sin(2a)\)

From this, we obtain:

\(\displaystyle a+\frac{\pi}{12}=2a\)

\(\displaystyle a=\frac{\pi}{12}\)

(ii) \(\displaystyle \sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a)\)

From this we obtain:

\(\displaystyle \pi-\left(a+\frac{\pi}{12}\right)=2a\)

\(\displaystyle 3a=\frac{11\pi}{12}\)

\(\displaystyle a=\frac{11\pi}{36}\)

Now, we must also consider (because of the periodicity of the sine function):

(iii) \(\displaystyle \sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a-2\pi)\)

From this we obtain:

\(\displaystyle \pi-\left(a+\frac{\pi}{12}\right)=2a-2\pi\)

\(\displaystyle 3a=\frac{35\pi}{12}\)

\(\displaystyle a=\frac{35\pi}{36}\)

This solution is outside of the given interval.

In light of the identities $\sin(x+\pi)=-\sin(x)$ and $\sin(-x)=-\sin(x)$, we have 1 more case to consider:

(iv) \(\displaystyle \sin\left(a+\frac{\pi}{12}+\pi\right)=\sin(-2a)\)

From this, we obtain:

\(\displaystyle a+\frac{\pi}{12}+\pi=-2a\)

\(\displaystyle 3a=-\frac{13\pi}{12}\)

\(\displaystyle a=-\frac{13\pi}{36}\)

This solution is outside of the given interval.

All other solutions resulting from periodicity are outside the given interval.

Thus, the only two solutions in the given interval are:

\(\displaystyle a\in\left\{\frac{\pi}{12},\frac{11\pi}{36}\right\}\)
 

FAQ: Which values of $a$ satisfy a trigonometric equation in a given interval?

What is a trigonometric equation?

A trigonometric equation is an equation that contains one or more trigonometric functions (such as sine, cosine, tangent) of one or more variables. It can be solved for values of the variable that make the equation true.

What is an interval in a trigonometric equation?

An interval in a trigonometric equation is a range of values for the variable that is being solved for. It can be specified using inequality symbols, such as x < 5 or 2 < x < 7.

How do I determine the values of a that satisfy a trigonometric equation in a given interval?

To determine the values of a that satisfy a trigonometric equation in a given interval, you can use algebraic techniques, such as factoring and substitution, to rearrange the equation and isolate the variable. Then, you can use the properties of trigonometric functions to find the values of a that make the equation true within the specified interval.

Can a trigonometric equation have multiple solutions for a given interval?

Yes, a trigonometric equation can have multiple solutions for a given interval. This is because trigonometric functions have periodic behavior, meaning that they repeat themselves at regular intervals. Therefore, there may be more than one value of a that satisfies the equation within a given interval.

How can I check my solutions for a trigonometric equation in a given interval?

You can check your solutions for a trigonometric equation in a given interval by plugging them back into the original equation and seeing if they make the equation true. You can also use a graphing calculator or software to visualize the graph of the equation and see if the solutions fall within the specified interval.

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