From the unit circle, we know that

[latex]\begin{gathered}\cos \theta =\frac{1}{2} \\ \theta =\frac{\pi }{3},\frac{5\pi }{3} \end{gathered}[/latex]

These are the solutions in the interval [latex]\left[0,2\pi \right][/latex]. All possible solutions are given by

[latex]\theta =\frac{\pi }{3}\pm 2k\pi \text{ and }\theta =\frac{5\pi }{3}\pm 2k\pi[/latex]

where [latex]k[/latex] is an integer.

Solving for all possible values of t means that solutions include angles beyond the period of [latex]2\pi[/latex]. From the unit circle, we can see that the solutions are [latex]t=\frac{\pi }{6}[/latex] and [latex]t=\frac{5\pi }{6}[/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

[latex]t=\frac{\pi }{6}\pm 2\pi k\text{ and }t=\frac{5\pi }{6}\pm 2\pi k[/latex]

where [latex]k[/latex] is an integer.

[latex]\begin{gathered}2\cos \theta -3=-5 \\ \cos \theta =-2 \\ \cos \theta =-1 \\ \theta =\pi \end{gathered}[/latex]

Make sure mode is set to radians. To find [latex]\theta[/latex], use the inverse sine function. On most calculators, you will need to push the 2^ND button and then the SIN button to bring up the [latex]{\sin }^{-1}[/latex] function. What is shown on the screen is [latex]{\sin}^{-1}[/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\sin }^{-1}\left(0.8\right)[/latex], and press ENTER. Thus, to four decimals places,

[latex]{\sin }^{-1}\left(0.8\right)\approx 0.9273[/latex]

This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]/pi - 0.9273 \approx 2.2143[/latex]

The general solution is

[latex]\theta \approx 0.9273\pm 2\pi k \text{ and } \theta \approx 2.2143 \pm 2\pi k[/latex]

The angle measurement in degrees is

[latex]\begin{align} \theta &\approx {53.1}^{\circ } \\ \theta &\approx {180}^{\circ }-{53.1}^{\circ } \\ &\approx {126.9}^{\circ } \end{align}[/latex]

Analysis of the Solution

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\pi -\theta[/latex].

We can begin with some algebra.

[latex]\begin{gathered}\sec \theta =-4\\ \frac{1}{\cos \theta }=-4\\ \cos \theta =-\frac{1}{4}\end{gathered}[/latex]

Check that the MODE is in radians. Now use the inverse cosine function.

[latex]\begin{gathered}{\cos }^{-1}\left(-\frac{1}{4}\right)\approx 1.8235 \\ \theta \approx 1.8235+2\pi k \end{gathered}[/latex]

Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14[/latex], 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.

Figure 2.

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta \text{ }\text{ }\text{'}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\theta \text{ }\text{ }\text{'}\approx \pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}[/latex]

The solutions are [latex]\theta \approx 1.8235\pm 2\pi k[/latex] and [latex]\theta \approx 4.4597\pm 2\pi k[/latex].

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\sin \theta[/latex]. Then we will find the angles.

[latex]\begin{gathered}2{\sin }^{2}\theta -1=0 \\ 2{\sin }^{2}\theta =1 \\ {\sin }^{2}\theta =\frac{1}{2} \\ \sqrt{{\sin }^{2}\theta }=\pm \sqrt{\frac{1}{2}} \\ \sin \theta =\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2} \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} \end{gathered}[/latex]

We want all values of [latex]\theta[/latex] for which [latex]\csc \theta =-2[/latex] over the interval [latex]0\le \theta <4\pi[/latex].

[latex]\begin{gathered}\csc \theta =-2 \\ \frac{1}{\sin \theta }=-2 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6},\frac{19\pi }{6},\frac{23\pi }{6} \end{gathered}[/latex]

Analysis of the Solution

As [latex]\sin \theta =-\frac{1}{2}[/latex], notice that all four solutions are in the third and fourth quadrants.

Recall that the tangent function has a period of [latex]\pi[/latex]. On the interval [latex]\left[0,\pi \right)[/latex], and at the angle of [latex]\frac{\pi }{4}[/latex], the tangent has a value of 1. However, the angle we want is [latex]\left(\theta -\frac{\pi }{2}\right)[/latex]. Thus, if [latex]\tan \left(\frac{\pi }{4}\right)=1[/latex], then

[latex]\begin{gathered}\theta -\frac{\pi }{2}=\frac{\pi }{4}\\ \theta =\frac{3\pi }{4}\pm k\pi \end{gathered}[/latex]

Over the interval [latex]\left[0,2\pi \right)[/latex], we have two solutions:

[latex]\theta =\frac{3\pi }{4}\text{ and }\theta =\frac{3\pi }{4}+\pi =\frac{7\pi }{4}[/latex]

We can solve this equation using only algebra. Isolate the expression [latex]\tan x[/latex] on the left side of the equals sign.

[latex]\begin{gathered} 2\left(\tan x\right)+2\left(3\right) =5+\tan x \\ 2\tan x+6 =5+\tan x \\ 2\tan x-\tan x =5 - 6 \\ \tan x =-1\end{gathered}[/latex]

There are two angles on the unit circle that have a tangent value of [latex]-1:\theta =\frac{3\pi }{4}[/latex] and [latex]\theta =\frac{7\pi }{4}[/latex].

We begin by using substitution and replacing cos [latex]\theta[/latex] with [latex]x[/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\cos \theta =x[/latex]. We have

[latex]{x}^{2}+3x - 1=0[/latex]

The equation cannot be factored, so we will use the quadratic formula [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].

[latex]\begin{align} x&=\frac{-3\pm \sqrt{{\left(-3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}&=\frac{-3\pm \sqrt{13}}{2} \end{align}[/latex]

Replace [latex]x[/latex] with [latex]\cos \theta[/latex], and solve. Thus,

[latex]\begin{gathered} \cos \theta =\frac{-3\pm \sqrt{13}}{2}\theta ={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\end{gathered}[/latex]

Note that only the + sign is used. This is because we get an error when we solve [latex]\theta ={\cos }^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)[/latex] on a calculator, since the domain of the inverse cosine function is [latex]\left[-1,1\right][/latex]. However, there is a second solution:

[latex]\begin{align}\theta &={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 1.26 \end{align}[/latex]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

[latex]\begin{align}\theta &=2\pi -{\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 5.02 \end{align}[/latex]

Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\sin \theta =u[/latex], or imagine it, as we factor:

[latex]\begin{gathered}2{\sin }^{2}\theta -5\sin \theta +3=0 \\ \left(2\sin \theta -3\right)\left(\sin \theta -1\right)=0 \end{gathered}[/latex]

Now set each factor equal to zero.

[latex]\begin{gathered}2\sin \theta -3=0 \\ 2\sin \theta =3 \\ \sin \theta =\frac{3}{2} \\ \text{ } \\ \sin \theta -1=0 \\ \sin \theta =1 \end{gathered}[/latex]

Next solve for [latex]\theta :\sin \theta \ne \frac{3}{2}[/latex], as the range of the sine function is [latex]\left[-1,1\right][/latex]. However, [latex]\sin \theta =1[/latex], giving the solution [latex]\theta =\frac{\pi }{2}[/latex].

Analysis of the Solution

Make sure to check all solutions on the given domain as some factors have no solution.

This problem should appear familiar as it is similar to a quadratic. Let [latex]\sin \theta =x[/latex]. The equation becomes [latex]2{x}^{2}+x=0[/latex]. We begin by factoring:

[latex]\begin{gathered}2{x}^{2}+x=0\\ x\left(2x+1\right)=0\end{gathered}[/latex]

Set each factor equal to zero.

[latex]\begin{gathered}x=0 \\ 2x+1=0 \\ x=-\frac{1}{2} \end{gathered}[/latex]

Then, substitute back into the equation the original expression [latex]\sin \theta[/latex] for [latex]x[/latex]. Thus,

[latex]\begin{gathered}\sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

The solutions within the domain [latex]0\le \theta <2\pi[/latex] are [latex]\theta =0,\pi ,\frac{7\pi }{6},\frac{11\pi }{6}[/latex]. If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[latex]\begin{gathered}2{\sin }^{2}\theta +\sin \theta =0 \\ \sin \theta \left(2\sin \theta +1\right)=0 \\ \sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ 2\sin \theta +1=0 \\ 2\sin \theta =-1 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

Analysis of the Solution

We can see the solutions on the graph in Figure 3. On the interval [latex]0\le \theta <2\pi[/latex], the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Figure 3

We can verify the solutions on the unit circle in Sum and Difference Identities as well.

We can factor using grouping. Solution values of [latex]\theta[/latex] can be found on the unit circle:

[latex]\begin{gathered}\left(2\sin \theta -1\right)\left(\sin \theta -1\right)=0 \\ 2\sin \theta -1=0 \\ \sin \theta =\frac{1}{2} \\ \theta =\frac{\pi }{6},\frac{5\pi }{6} \\ \text{ } \\ \sin \theta =1 \\ \theta =\frac{\pi }{2} \end{gathered}[/latex]

Notice that the left side of the equation is the difference formula for cosine.

[latex]\begin{align} \cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)&=\frac{\sqrt{3}}{2} \\ \cos \left(x - 2x\right)&=\frac{\sqrt{3}}{2}&& \text{Difference formula for cosine} \\ \cos \left(-x\right)&=\frac{\sqrt{3}}{2}&& \text{Use the negative angle identity}. \\ \cos x&=\frac{\sqrt{3}}{2}\end{align}[/latex]

From the unit circle in Sum and Difference Identities, we see that [latex]\cos x=\frac{\sqrt{3}}{2}[/latex] when [latex]x=\frac{\pi }{6},\frac{11\pi }{6}[/latex].

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

[latex]\begin{gathered}\cos \left(2\theta \right)=\cos \theta \\ 2{\cos }^{2}\theta -1=\cos \theta \\ 2{\cos }^{2}\theta -\cos \theta -1=0 \\ \left(2\cos \theta +1\right)\left(\cos \theta -1\right)=0 \\ 2\cos \theta +1=0 \\ \cos \theta =-\frac{1}{2} \\ \text{ } \\ \cos \theta -1=0 \\ \cos \theta =1 \end{gathered}[/latex]

So, if [latex]\cos \theta =-\frac{1}{2}[/latex], then [latex]\theta =\frac{2\pi }{3}\pm 2\pi k[/latex] and [latex]\theta =\frac{4\pi }{3}\pm 2\pi k[/latex]; if [latex]\cos \theta =1[/latex], then [latex]\theta =0\pm 2\pi k[/latex].

If we rewrite the right side, we can write the equation in terms of cosine:

[latex]\begin{align} 3 \cos\theta +3& =2 \sin^{2}\theta \\ 3 \cos\theta +3& =2\left(1-{\cos}^{2}\theta \right) \\ 3 \cos\theta +3& =2 - 2\cos^{2}\theta \\ 2{\cos }^{2}\theta +3 \cos\theta +1& =0 \\ \left(2 \cos\theta +1\right)\left(\cos \theta +1\right)& =0 \\ 2 \cos\theta +1& =0 \\ \cos \theta & =-\frac{1}{2} \\ \theta & =\frac{2\pi }{3},\frac{4\pi }{3} \\ \text{ } \\ \cos \theta +1& =0 \\ \cos \theta & =-1 \\ \theta & =\pi\end{align}[/latex]

Our solutions are [latex]\theta =\frac{2\pi }{3},\frac{4\pi }{3},\pi[/latex].

We can see that this equation is the standard equation with a multiple of an angle. If [latex]\cos \left(\alpha \right)=\frac{1}{2}[/latex], we know [latex]\alpha[/latex] is in quadrants I and IV. While [latex]\theta ={\cos }^{-1}\frac{1}{2}[/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\cos \theta =\frac{1}{2}[/latex] will be in quadrants I and IV.

Therefore, the possible angles are [latex]\theta =\frac{\pi }{3}[/latex] and [latex]\theta =\frac{5\pi }{3}[/latex]. So, [latex]2x=\frac{\pi }{3}[/latex] or [latex]2x=\frac{5\pi }{3}[/latex], which means that [latex]x=\frac{\pi }{6}[/latex] or [latex]x=\frac{5\pi }{6}[/latex]. Does this make sense? Yes, because [latex]\cos \left(2\left(\frac{\pi }{6}\right)\right)=\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}[/latex].

Are there any other possible answers? Let us return to our first step.

In quadrant I, [latex]2x=\frac{\pi }{3}[/latex], so [latex]x=\frac{\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align} 2x&=\frac{\pi }{3}+2\pi \\ &=\frac{\pi }{3}+\frac{6\pi }{3} \\ &=\frac{7\pi }{3} \end{align}[/latex]

so [latex]x=\frac{7\pi }{6}[/latex].

One more rotation yields

[latex]\begin{align} 2x&=\frac{\pi }{3}+4\pi \\ &=\frac{\pi }{3}+\frac{12\pi }{3} \\ &=\frac{13\pi }{3} \end{align}[/latex]

[latex]x=\frac{13\pi }{6}>2\pi[/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi[/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

In quadrant IV, [latex]2x=\frac{5\pi }{3}[/latex], so [latex]x=\frac{5\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align}2x&=\frac{5\pi }{3}+2\pi \\ &=\frac{5\pi }{3}+\frac{6\pi }{3} \\ &=\frac{11\pi }{3}\end{align}[/latex]

so [latex]x=\frac{11\pi }{6}[/latex].

One more rotation yields

[latex]\begin{align}2x&=\frac{5\pi }{3}+4\pi \\ &=\frac{5\pi }{3}+\frac{12\pi }{3} \\ &=\frac{17\pi }{3} \end{align}[/latex]

[latex]x=\frac{17\pi }{6}>2\pi[/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi[/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

Our solutions are [latex]x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\text{and }\frac{11\pi }{6}[/latex]. Note that whenever we solve a problem in the form of [latex]\sin \left(nx\right)=c[/latex], we must go around the unit circle [latex]n[/latex] times.

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

[latex]\begin{gathered}{a}^{2}+{b}^{2}={c}^{2} \\ {\left(23\right)}^{2}+{\left(69.5\right)}^{2}\approx 5359 \\ \sqrt{5359}\approx 73.2\text{ m}\end{gathered}[/latex]

The angle of elevation is [latex]\theta[/latex], formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

[latex]\begin{align}\tan \theta &=\frac{69.5}{23} \\ {\tan }^{-1}\left(\frac{69.5}{23}\right)&\approx 1.2522 \\ &\approx {71.69}^{\circ } \end{align}[/latex]

The angle of elevation is approximately [latex]{71.7}^{\circ }[/latex], and the length of the cable is 73.2 meters.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet.

Figure 5

The side adjacent to [latex]\theta[/latex] is a and the hypotenuse is [latex]4a[/latex]. Thus,

[latex]\begin{gathered}\cos \theta =\frac{a}{4a}=\frac{1}{4} \\ {\cos }^{-1}\left(\frac{1}{4}\right)\approx {75.5}^{\circ } \end{gathered}[/latex]

The elevation of the ladder forms an angle of [latex]{75.5}^{\circ }[/latex] with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

[latex]\begin{gathered}{a}^{2}+{b}^{2}={\left(4a\right)}^{2} \\ {b}^{2}={\left(4a\right)}^{2}-{a}^{2} \\ {b}^{2}=16{a}^{2}-{a}^{2} \\ {b}^{2}=15{a}^{2} \\ b=\sqrt{15}a \end{gathered}[/latex]

Thus, the ladder touches the wall at [latex]\sqrt{15}a[/latex] feet from the ground.