Solving Trigonometric Equations

Now, try to find the angle $\theta$ for which the value of $\tan \theta$ equals $1.$ The table of the trigonometric values of notable angles can be useful here. It can be concluded that this equation has two possible solutions. There are two different trigonometric functions involved in the equation, sine and cosine. Therefore, the first step is to rewrite one side of the equation in terms of the other. To do this, the Double-Angle Identity for cosine can be used. Next, to solve this equation, rewrite the left-hand side as a product of two expressions. By the Zero Product Property, in order for the equation to be true, at least one of the parentheses should be equal to $0.$ Note that adding $7$ to both sides of the Equation $(\text{II})$ results in $\sin \theta =7.$ However, this is not possible, as the values of sine are always between $\text{-}1$ and $1.$ Therefore, this equation has no solution. Now consider Equation $(\text{I}).$ To find the value of ${\sin }^{\text{-}1}\left(\frac{1}{2}\right),$ or arcsine of $\frac{1}{2},$ think about the angles for which the value of sine is $\frac{1}{2}.$ According to the table of the trigonometric ratios of common angles, there are two such angles in the interval between $0^{\circ }$ and $360^{\circ }.$ Therefore, the given equation has two possible solutions. Now that all the solutions have been found, the code mentioned at the beginning can be determined. In Part A, the solutions $45$ and $225$ were found, while the solutions from Part B were $30$ and $150.$ The code can be found by combining these numbers in the order directed at the start. Try it out to see the hidden message!

Apply this identity and then simplify the equation. In order to find the values of $\theta$ that satisfy this equation, think about the angles for which the cosine equals $\text{-}1.$ According to the table of the trigonometric ratios of common angles, there is at least one such angle. Since cosine is a periodic function with a period of $2\pi$, for every integer $n,$ the following angles are the solutions to the equation. There are two trigonometric functions in the equation, tangent and secant. Recall that one of the Pythagorean Identities relates these two functions. Substitute $1+{\tan }^2\theta$ for ${\sec }^2\theta$ into the equation and simplify it. Next, look for the angles for which tangent is equal to $1$ or $\text{-}1.$ To find the general equation for the solutions, note that all these angles can be expressed in the following way. Therefore, all the solutions to the equation can be given by the following equation where $n$ is an integer number. After correctly solving both equations, Magdalena can finally read the joke that Davontay found for her.

Since there are two different trigonometric functions, one should be rewritten in terms of the other. To do so, use the Pythagorean Identity, but first rewrite the equation so that ${\cos }^2\theta$ is isolated on one side of the equation. Substitute $1-{\sin }^2\theta$ for ${\cos }^2\theta$ and simplify the equation. Next, factor the expression on the left-hand side into two parentheses. By the Zero Product Property, at least one of the parentheses must be equal to $0.$ Two equations can be formed by applying this property. To solve these two equations, try to find the angles for which the values of sine are $\frac{1}{2}$ and $1.$ For the first equation, there are two such angles — $\frac{\pi }{6}$ and $\frac{5\pi }{6}$ — while for the second there is only one — $\frac{\pi }{2}.$ Finally, the solutions should be checked whether there are any extraneous solutions. To do that, substitute each of them into the original equation and see if it is true. The other two solutions can be checked in a similar fashion.

Solution	Substitute	Evaluate	True or False
$\theta =\frac{\pi }{6}$	$2{\cos }^2\left(\frac{\pi }{6}\right)+3\sin \left(\frac{\pi }{6}\right)\stackrel{?}{=}3$	$2(\frac{\sqrt{3}}{2}{)}^2+3(\frac{1}{2})\stackrel{?}{=}3$	$3=3✓$
$\theta =\frac{5\pi }{6}$	$2{\cos }^2\left(\frac{5\pi }{6}\right)+3\sin \left(\frac{5\pi }{6}\right)\stackrel{?}{=}3$	$2(\text{-}\frac{\sqrt{3}}{2}{)}^2+3(\frac{1}{2})\stackrel{?}{=}3$	$3=3✓$
$\theta =\frac{\pi }{2}$	$2{\cos }^2\left(\frac{\pi }{2}\right)+3\sin \left(\frac{\pi }{2}\right)\stackrel{?}{=}3$	$2(0{)}^2+3(1)\stackrel{?}{=}3$	$3=3✓$

Therefore, there are three solutions to the equation and none of them are extraneous.

First, move $\cos \theta$ to the right-hand side of the equation. Then, square each side of the equation. Now, to rewrite one trigonometric function in terms of the other, use the Pythagorean Identity. Similarly to Part A, substitute $1-{\sin }^2\theta$ for ${\cos }^2\theta .$ To find the solutions of this equation, look for the angles for which the sine equals $\pm \frac{\sqrt{2}}{2}.$ The table with the trigonometric ratios of notable angles can be useful here. Finally, substitute each of these solutions into the original equation in order to determine whether any of them is extraneous.

Solution	Substitute	Evaluate	True or False
${\theta }_1=\frac{\pi }{4}$	$\sin \left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{4}\right)\stackrel{?}{=}0$	$\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\stackrel{?}{=}0$	$0=0✓$
${\theta }_2=\frac{3\pi }{4}$	$\sin \left(\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)\stackrel{?}{=}0$	$\frac{\sqrt{2}}{2}-\left(\text{-}\frac{\sqrt{2}}{2}\right)\stackrel{?}{=}0$	$\sqrt{2}\ne 0\times$
${\theta }_3=\frac{5\pi }{4}$	$\sin \left(\frac{5\pi }{4}\right)-\cos \left(\frac{5\pi }{4}\right)\stackrel{?}{=}0$	$\text{-}\frac{\sqrt{2}}{2}-\left(\text{-}\frac{\sqrt{2}}{2}\right)\stackrel{?}{=}0$	$0=0✓$
${\theta }_4=\frac{7\pi }{4}$	$\sin \left(\frac{7\pi }{4}\right)-\cos \left(\frac{7\pi }{4}\right)\stackrel{?}{=}0$	$\text{-}\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\stackrel{?}{=}0$	$\text{-}\sqrt{2}\ne 0\times$

It can be concluded that $\theta =\frac{3\pi }{4}$ and $\theta =\frac{7\pi }{4}$ are extraneous solutions. Therefore, only $\theta =\frac{\pi }{4}$ and $\theta =\frac{5\pi }{4}$ are solutions to the equation. This means that Davontay bet on the right equation and he will get the last piece of the cake!

When $\sin \theta =0,$ the cotangent of $\theta$ is undefined, as division by zero cannot be calculated. According to the table of trigonometric values of common angles, $\sin \theta$ equals $0$ for $\theta$ equal $\pi ,$ $2\pi ,$ $3\pi ,$ and so on. All these angles can be described by the expression $\pi n,$ where $n$ is an integer number. Therefore, these values of $\theta$ cannot be solutions to the equation. Now, substitute $\frac{\cos \theta }{\sin \theta }$ for $\cot \theta$ and simplify the equation. By the Zero Product Property, at least one factor in the equation should be equal to $0.$ From this point, two equations can be formed. The solutions of the equations can be found by identifying the angles for which cosine is $0$ and $1.$ Again, using the table of trigonometric values of common angles, it can be found that there are two solutions to Equation (I), $\frac{\pi }{2}$ and $\frac{3\pi }{2}.$ Since cosine is a periodic function, these two solutions repeat every period of $2\pi .$ The solutions can be visualized on the following diagram. These solutions together can be described by the following general equation. Similarly, the table of trigonometric values can be used to find that Equation (II) has one solution, $0,$ which also repeats every period of $2\pi .$ However, recall that $\theta =\pi n$ cannot be a solution to the equation because cotangent is undefined for these values. This means that the equation itself is undefined for these values. Therefore, only ${\theta }_1=\frac{\pi }{2}+\pi n$ are solutions to the initial equation. Finally, graph these solutions on a unit circle to see which directions are indicated.

As shown on the unit circle, the solutions are located at two out of the four cardinal directions, north and south. Therefore, these are the directions the students should not choose.

According to this identity, $\cos \theta$ can be substituted for $\cos (\text{-}\theta )$ on the right-hand side of the equation. Additionally, one of the Pythagorean Identities that involves secant can be used to simplify the left-hand side of the equation. To find the solutions to this equation, use the table of trigonometric ratios of notable angles again. According to the table, $\cos \theta =\text{-}1$ for $\theta =\pi .$ Furthermore, since cosine is a periodic function, it has the same value for the angles $\pi +2\pi ,$ $\pi +4\pi ,$ and so on. Therefore, all the solutions can be described by the following equation. Finally, plot the solution on a unit circle.

The solution is located in the western direction, which means that students should not choose it. Considering the solutions to the first equation, the only direction left is east, so the students should turn east at the crossroad.

Use this identity to rewrite the equation and then simplify it. Two equations can be formed by applying the Zero Product Property. Finally, try to identify angles for which the values of sine is $\frac{1}{2}$ or the values of cosine is $\frac{\sqrt{3}}{2}.$ Here, ${\theta }_1$ and ${\theta }_2$ are the solutions to Equation (I), while ${\theta }_3$ and ${\theta }_4$ are the solutions to Equation (II). Since $0\le \theta \le 360$ and two of the angles are the same, there are three solutions to the equation. Now, look for the angles for which sine is equal to $1$ or $\text{-}1.$ Here the table of trigonometric ratios of notable angles can be useful. As can be seen, there are two solutions to the equation. Therefore, the first equation has more solutions.

In order to calculate when the seat would be $34$ meters above the ground, substitute $34$ for $h$ and solve equation for $t.$ To solve this equation, use the inverse function of cosine — arccosine. Apply it to the both sides of the equation. Since the value of $\arccos \left(\text{-}\frac{1}{2}\right)$ is known, the equation can be further simplified until the solution is found. Next, the value of ${\cos }^{\text{-}1}\left(\text{-}\frac{1}{2}\right)$ can be found by looking for an angle for which the value of cosine is $\text{-}\frac{1}{2}.$ By using the table of the values of notable angle, it can be concluded that there are two such angles, $\frac{2\pi }{3}$ and $\frac{4\pi }{3}.$ What is more, since cosine is a periodic functions, those angles will appear every $2\pi$ periods. This fact can be represented by adding $2\pi k$ to the found solutions where $k$ is an integer number. Finally, divide both sides of the equations by $3\pi$ to find the values of $t.$ The girls want to find when the seat will be $34$ meters above the ground for the first time. Therefore, the least value of $t$ should be found. Note that when $k=0,$ $t_1$ has the smallest value out of all. It can be concluded that they will be $34$ meters above the ground after $\frac{2}{9}$ of a minute. Use a conversion factor of $\frac{60\text{ sec}}{1\text{ min}}$ to rewrite $\frac{2}{9}$ minutes in seconds. Therefore, Magdalena and Paulina will be $34$ meters high after only $13.3$ seconds of riding on the Ferris wheel. The seat will be $23$ meters above the ground after $3.5$ minutes of riding.