Algebra Refresher
Solve.
- [latex]\displaystyle (2x-6)^2=4[/latex]
- [latex]\displaystyle \sqrt{2x-6}=4[/latex]
- [latex]\displaystyle \dfrac{1}{2x-6}=4[/latex]
- [latex]\displaystyle \lvert {2x-6} \rvert =4[/latex]
- [latex]\displaystyle \log{2x-6}=4[/latex]
- [latex]\displaystyle 2^{2x-6}=4[/latex]
Algebra Refresher Answers
- [latex]\displaystyle 2,4[/latex]
- [latex]\displaystyle 11[/latex]
- [latex]\displaystyle \dfrac{25}{8}[/latex]
- [latex]\displaystyle 1,5[/latex]
- [latex]\displaystyle 5003[/latex]
- [latex]\displaystyle 4[/latex]
- Find exact solutions to equations of the form [latex]\sin nx = k[/latex]
- Find all solutions between [latex]0[/latex] and [latex]2\pi[/latex]
- Use a substitution to solve trigonometric equations
- Write expressions for exact solutions
- Solve problems involving trigonometric models
We have seen that an equation of the form [latex]\sin \theta = k[/latex] (for [latex]-1 \lt k \lt 1[/latex]) always has two solutions between [latex]0[/latex] and [latex]2\pi {.}[/latex] For example, the figure below illustrates that [latex]x=\dfrac{\pi}{6}[/latex] and [latex]x=\dfrac{5\pi}{6}[/latex] are solutions of the equation [latex]\sin x = 0.5{.}[/latex] These two solutions correspond to the two points on the unit circle where [latex]y=0.5{.}[/latex]
The calculator gives us only one of these solutions, but we can use reference angles to find the other solution. In fact, if we use a calculator to find one solution as [latex]\theta_{1}=\sin^{-1}k{,}[/latex] then the other solution is [latex]\theta_{2}=\pi - \theta_{1}{.}[/latex] You can see this by considering the symmetry of the \sine graph, or of the unit circle, as shown below.
This relationship between the two solutions still holds if [latex]k[/latex] is negative, because the calculator returns a negative angle for [latex]\theta_{1}=\sin^{-1}k{.}[/latex] See the figure below.
Solve [latex]~~\sin \theta = -0.58~~[/latex] for [latex]0 le \theta \lt 2\pi{.}[/latex]
Solution
The calculator gives us the solution
[latex]\sin^{-1}(-0.58)=-0.6187[/latex]
However, because this solution is not between [latex]0[/latex] and [latex]2\pi{,}[/latex] we find a coterminal angle,
[latex]\theta_{1}=\sin^{-1}(-0.58)+2\pi = -0.6187+2\pi=5.6645[/latex]
for the solution in the fourth quadrant. The second quadrant solution is
[latex]\theta_2=\pi-\sin^{-1}(-0.58)=\pi +0.6187=3.7603[/latex]
To four decimal places, the desired solutions are 3.7603 and 5.6645.
A similar method applies to equations of the form [latex]\cos \theta = k{.}[/latex] The two solutions between [latex]0[/latex] and [latex]2\pi[/latex] are [latex]\theta_{1}=\cos^{-1}(k)[/latex] and [latex]\theta_{2}=2\pi -\cos^{-1}(k){.}[/latex] Once again, you can see this by considering the symmetry of the co\sine graph, or of the unit circle, as shown below.
The relationship between the two solutions still holds if [latex]k[/latex] is negative. See the figure below.
Checkpoint 7.27.
Solve [latex]~~\cos \theta = -0.36~~[/latex] for [latex]0 le \theta \lt 2\pi{.}[/latex]
Solution
1.9391, 4,3441
Equations of the form [latex]tan \theta = k[/latex] are generally easier to solve, because there is only one solution in each cycle of the graph. Each solution differs from the previous one by [latex]\pi{,}[/latex] as shown below.
For example, to solve the equation [latex]tan \theta = -2.4[/latex] for [latex]0 le \theta \lt 2\pi{,}[/latex] we first calculate
[latex]tan^{-1}(-2.4)=-1.1760[/latex]
Because this angle is not between [latex]0[/latex] and [latex]2\pi{,}[/latex] we add [latex]\pi[/latex] to find the next two solutions:
[latex]\theta_{1} = -1.1760+\pi=1.9656\ \theta_{2} = 1.9656+\pi=5.1072[/latex]
We summarize these observations for the three trigonometric functions as follows.
- The equation [latex]\cos \theta = k, ~~ -1\lt k \lt 1{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]\theta_{1}=\cos^{-1}(k)~~ {and}~~ \theta_{2}=2\pi -\theta_{1}[/latex]
- The equation [latex]\sin \theta = k, ~~ -1\lt k \lt 1{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]{If}~k \gt 0:~~ \theta_{1} =\sin^{-1}(k)~~ {and}~~ \theta_{2}=\pi -\theta_{1}[/latex][latex]{If}~k \lt 0:~~ \theta_{1}=\sin^{-1}(k) + 2\pi ~~ {and}~~ \theta_{2}=\pi -\sin^{-1}(k)[/latex]
- The equation [latex]tan \theta = k{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]{If}~k \gt 0:~~ \theta_{1}=tan^{-1}(k)~~ {and}~~ \theta_{2}=\pi +\theta_{1}[/latex][latex]{If}~k \lt 0:~~ \theta_{1}=tan^{-1}(k) + \pi ~~ {and}~~ \theta_{2}=\pi +\theta_{1}[/latex]
If [latex]n[/latex] is an integer, what can we say about the solutions of the equation
[latex]\sin nx = k~~ {?}[/latex]
For example, how many solutions are there for the equation [latex]\sin 2x=0.5{?}[/latex] The figure below shows that this equation has four solutions between [latex]0[/latex] and [latex]2\pi{.}[/latex] The graph of [latex]y=\sin 2x[/latex] completes two cycles between [latex]0[/latex] and [latex]2\pi{,}[/latex] and each cycle produces two solutions, for a total of four.
Find all solutions of [latex]~~\sin 2x=0.5~~[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
We begin as usual, by taking the inverse sine of each side of the equation, to get
[latex]2x=\sin^{-1}(0.5)[/latex]
There are two values of [latex]2x[/latex] between [latex]0[/latex] and [latex]2\pi[/latex] with [latex]\sin 2x=0.5[/latex]; namely, [latex]2x=\dfrac{\pi}{6}[/latex] and [latex]2x=\pi - \dfrac{\pi}{6}=\dfrac{5\pi}{6}{.}[/latex] Solving these equations for [latex]x[/latex] yields [latex]x=\dfrac{\pi}{12}[/latex] and [latex]x=\dfrac{5\pi}{12}{.}[/latex]
But these are only two of the four solutions! The graph of [latex]y=\sin 2x[/latex] completes two cycles between [latex]0[/latex] and [latex]2\pi{,}[/latex] each of length [latex]\pi{.}[/latex] We must add [latex]\pi[/latex] to each of the first two solutions to find the solutions in the second cycle. These two solutions are
[latex]\dfrac{\pi}{12}+\pi=\dfrac{13\pi}{12}~~ {and}~~ \dfrac{5\pi}{12}+\pi=\dfrac{17\pi}{12}[/latex]
Note that the solutions in the second cycle are still less than [latex]2\pi{,}[/latex] so they must be included in the set of all solutions between [latex]0[/latex] and [latex]2\pi{.}[/latex] The four solutions are [latex]\dfrac{\pi}{12},~\dfrac{5\pi}{12},~\dfrac{13\pi}{12}[/latex] and [latex]\dfrac{17\pi}{12}{.}[/latex]
Checkpoint 7.29.
- Sketch a graph of [latex]~~y=\cos 2x~~[/latex] for [latex]0 \le \theta \le 2\pi{.}[/latex]
- Find exact values for all solutions of [latex]\cos 2x=\dfrac{-\sqrt{2}}{2}[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
- [latex]\displaystyle \dfrac{3\pi}{8},~\dfrac{5\pi}{8},~\dfrac{11\pi}{8},~\dfrac{13\pi}{8}[/latex]
As we observed earlier, equations involving the tangent function are easier to solve, because there is only one solution in each cycle of the graph. Once we have found one solution, we can find all the others by adding multiples of the period.
Find all solutions of [latex]~~\tan 2t=-1~~[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
We expect to find four solutions between [latex]0[/latex] and [latex]2\pi{.}[/latex] The first angle [latex]2t[/latex] between [latex]0[/latex] and [latex]2\pi[/latex] whose tangent is 1 is [latex]2t=\dfrac{3\pi}{4}{,}[/latex] so the first solution is [latex]t_{1}=\dfrac{3\pi}{8}{.}[/latex] Now, the period of the graph of [latex]y=\tan 2t[/latex] is [latex]\dfrac{\pi}{2}{,}[/latex] as shown at right.
There is one solution in each cycle, so we add multiples of [latex]\dfrac{\pi}{2}[/latex] to [latex]t_{1}[/latex] to find the other solutions.
[latex]t_{2}=\dfrac{3\pi}{8}+\dfrac{\pi}{2}=\dfrac{7\pi}{8}\ t_{3}=\dfrac{7\pi}{8}+\dfrac{\pi}{2}=\dfrac{11\pi}{8}\ t_{4}=\dfrac{11\pi}{8}+\dfrac{\pi}{2}=\dfrac{15\pi}{8}[/latex]
The four solutions are [latex]\dfrac{3\pi}{8},~ \dfrac{7\pi}{8},~ \dfrac{11\pi}{8}[/latex] and [latex]\dfrac{15\pi}{8}{.}[/latex]
Checkpoint 7.31.
Find all solutions of [latex]~~\tan 2x=\sqrt{3}~~[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
[latex]\dfrac{\pi}{6},~\dfrac{2\pi}{3},~\dfrac{7\pi}{6},~\dfrac{5\pi}{3}[/latex]
The larger the value of [latex]n{,}[/latex] the more cycles the graph completes between [latex]0[/latex] and [latex]2\pi{,}[/latex] and the more solutions we find. Thus, for [latex]-1 \lt k \lt 1{,}[/latex] there are six solutions of [latex]\sin 3x=k[/latex] between [latex]0[/latex] and [latex]2\pi{,}[/latex] eight solutions of [latex]\sin 4x=k{,}[/latex] and so on. (See the figure below.)
Of course, if [latex]k[/latex] is not one of the special values, we’ll need a calculator to help us solve the equation.
Solve [latex]~~1+4\cos 2x=2~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
Solution
We first solve for the trig ratio, [latex]\cos 2x[/latex]
[latex]1+4\cos 2x = 2[/latex] Subtract 1 from both sides.
[latex]4\cos 2x = 1[/latex] Divide both sides by 4.
[latex]\cos 2x = 0.25[/latex]
Next, we use a calculator to find two solutions of [latex]\cos 2x=0.25{.}[/latex] The solutions of this equation, rounded to three decimal places, are
[latex]2x = 1.318[/latex] and [latex]2x = 4.965[/latex]
[latex]x_{1} = 0.659[/latex] [latex]x_{2} =2.483[/latex]
Finally, because the period of the function [latex]f(x)=\cos 2x[/latex] is [latex]\pi{,}[/latex] we can find the other two solutions by adding [latex]\pi[/latex] to the first two solutions to get
[latex]x_{3}=0.659+\pi = 3.801\ x_{4}=2.483+\pi=5.624[/latex]
To two decimal places, the four solutions are 0.66, 2.48, 3.80, and 5.62
Checkpoint 7.33.
Solve [latex]~~5-2\sin 3x = 3.7~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
Solution
0.24, 1.86, 2.33, 2,95, 4.43, 6.06
For more complicated equations, it can be helpful to use a substitution in order to reduce the equation to the form [latex]\sin \theta=k[/latex] (or [latex]\cos \theta=k[/latex] or [latex]\tan \theta=k[/latex]). We want our substitution to replace the input of the sine function by a single variable. For example, in the next example, we substitute [latex]\theta[/latex] for the angle [latex]2x+1.5{,}[/latex] so that the equation [latex]\sin (2x+1.5)=-0.3[/latex] becomes [latex]\sin \theta = -0.3{.}[/latex]
Use a substitution to solve [latex]~~\sin (2x+1.5)=-0.3~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
Solution
We expect to find four solutions. Let [latex]\theta = 2x+1.5{,}[/latex] and use a calculator to find two solutions of [latex]\sin \theta = -0.3{.}[/latex] The solutions, rounded to three decimal places, are
[latex]\theta_{1}=\sin^{-1}(-0.3)=-0.3047 ~~ {and} ~~ \theta_{2}=\pi - \sin^{-1}(-0.3)=3.4463[/latex]
Because [latex]\theta[/latex] is not between [latex]0[/latex] and [latex]2\pi{,}[/latex] we find a coterminal angle,
[latex]\theta_{1}=\sin^{-1}(-0.3)+2\pi = 5.9785[/latex]
Next we replace [latex]\theta[/latex] by [latex]2x+1.5[/latex] to find two of the solutions of the original equation:
[latex]2x+1.5 = 3.4463 {and} 2x+1.5 = 5.9785\ x_{1} = 0.9731 x_{2} =2.2392[/latex]
Finally, because the period of the function [latex]f(x)=\sin (2x+1.5)[/latex] is [latex]\pi{,}[/latex] we find the other two solutions by adding [latex]\pi[/latex] to the first two solutions to get
[latex]x_{3}=0.9731+\pi = 4.1147\ x_{4}=2.2392+\pi=5.3808[/latex]
Rounded to hundredths, the four solutions are 0.97, 2.24, 4.11, and 5.38.
Checkpoint 7.35.
Use a substitution to solve [latex]~~4\cos (3x-0.5) = -3.2~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
Solution
[latex]1.00, 1.43, 3.09, 3.52, 5.19, 5.62[/latex]
Here is our strategy for solving trigonometric equations by using a substitution.
To solve the equation [latex]\sin (Bx+C)=k[/latex] or [latex]\cos (Bx+C)=k{:}[/latex]
-
- Substitute [latex]\theta = Bx+C{,}[/latex] and find two solutions for [latex]\sin \theta = k[/latex] or [latex]\cos \theta = k{.}[/latex]
2. Replace [latex]\theta[/latex] by [latex]Bx+C[/latex] in each solution and solve for [latex]x{.}[/latex]
3. Find the other solutions by adding multiples of [latex]\dfrac{2\pi}{B}[/latex] to the first two solutions.
To solve the equation [latex]\tan (Bx+C)=k{:}[/latex]
-
- Substitute [latex]\theta = Bx+C{,}[/latex] and find one solution for [latex]\tan \theta = k{.}[/latex]
2. Replace [latex]\theta[/latex] by [latex]Bx+C[/latex] and solve for [latex]x{.}[/latex]
3. Find the other solutions by adding multiples of [latex]\dfrac{\pi}{B}[/latex] to the first solution.
Find all solutions between [latex]0[/latex] and [latex]2\pi[/latex] to [latex]~~5+0.4\tan (3x-0.5)=4.5{.}[/latex]
Solution
The graph of [latex]f(x)=5+0.4\tan (3x-0.5)[/latex] completes six cycles between [latex]0[/latex] and [latex]2\pi{,}[/latex] so we expect to find six solutions, as illustrated below. We’ll use the substitution [latex]\theta=3x-0.5[/latex] to reduce the equation to [latex]5+0.4\tan \theta)=4.5{.}[/latex] Next, we isolate the trig ratio. Subtract 5 from both sides of the equation, then divide by 0.4.
[latex]0.4\tan \theta = -0.5\ \tan \theta = -1.25[/latex]
Now we can solve for [latex]\theta{.}[/latex]
[latex]\theta = \tan^{-1}(-1.25)=-0.8960[/latex]
Replacing [latex]\theta[/latex] by [latex]3x-0.5{,}[/latex] we find the first solution.
[latex]3x-0.5 = -0.8960\ 3x = -0.3960\ x = -0.1320[/latex]
This value of [latex]x[/latex] is not between [latex]0[/latex] and [latex]2\pi{,}[/latex] but because the period of [latex]f(x)[/latex] is [latex]\dfrac{\pi}{3}{,}[/latex] we can add [latex]\dfrac{\pi}{3} \approx 1.0472[/latex] to any solution to find another solution.
[latex]x_{1} =-0.1320+1.0472=0.9152\ x_{2} =0.9152+1.0472=1.9624\ x_{3} =1.9624+1.0472=3.0096\ x_{4} =3.0096+1.0472=4.0568\ x_{5} =4.0568+1.0472=5.1040\ x_{6} =5.1040+1.0472=6.1512[/latex]
We stop here, because the next solution is greater than [latex]2\pi{.}[/latex] Rounded to two decimal places, the six solutions are 0.92, 1.96, 3.01, 4.06, 5.10, and 6.15.
Checkpoint 7.37.
-
- Graph [latex]~~y=2-4\tan 3(x+0.2)~~[/latex] from [latex]x=0[/latex] to [latex]x=2\pi{.}[/latex]
Find all solutions of [latex]2-4\tan 3(x+0.2)=5[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex] Round your answers to two decimal places.
Solution
- [latex]\displaystyle 0.63, 1.68, 2,73, 3.77, 4.82, 5.87[/latex]
Trigonometric equations often arise in the study of periodic models.
A piston is pumping vertically at a rate of 1000 cycles per second. The distance between its lowest and highest position is 16 centimeters.
-
- Suppose the piston is at its midline position at [latex]t=0[/latex] and moving downward. Find a formula for the sinusoidal function [latex]h(t)[/latex] that gives the piston’s height.
- Use the formula to find the first two times the piston is 14 centimeters above its lowest position.
Solution
-
- Because the piston starts at its midline and moves down, we will write [latex]h(y)[/latex] as a sine function:[latex]h(t)=-A\sin (Bt) + k[/latex]The amplitude is [latex]A=\dfrac{16}{2} = 8{,}[/latex] and the midline is [latex]k=8{.}[/latex] The period is [latex]\dfrac{1}{1000}{,}[/latex] so [latex]\dfrac{B}{1000} = 2\pi{,}[/latex] and [latex]B=2000\pi{.}[/latex] Thus[latex]h(t)=-8\sin (2000\pi t) + 8[/latex]The graph is shown at right.
- With [latex]h(t)=14{,}[/latex] we have
[latex]-8\sin (2000\pi t) + 8 = 14[/latex] Subtract 8 from both sides.
[latex]- 8\sin (2000\pi t) = 6[/latex] Divide both sides by -8.
[latex]\sin (2000\pi t) =\dfrac{-3}{4}[/latex]
We use the substitution [latex]\theta = 2000\pi t[/latex] and find two positive solutions of the equation [latex]\sin \theta = \dfrac{-3}{4}{.}[/latex]
[latex]\theta = \sin^{-1}\left(\dfrac{-3}{4}\right)2\pi = 5.4351 ~~ {and} ~~ \theta = \pi - \sin^{-1}\left(\dfrac{-3}{4}\right)=3.9897[/latex]
Replace [latex]\theta[/latex] by [latex]2000\pi t[/latex] and solve each equation for [latex]t{:}[/latex]
[latex]2000\pi t = 5.4351 {and} 2000\pi t = 3.9897\ t = 0.000865 t =0.000635[/latex]
Thus, the first two times the piston is at a height of 14 centimeters are approximately [latex]t = 0.000865[/latex] and [latex]t =0.000635[/latex] seconds.
Checkpoint 7.39.
A lever on an oil well is pumping vertically at a rate of 10 cycles per minute. The distance between its lowest position at the ground and its highest position is 1.8 meters.
- Suppose the lever is at its midline position at [latex]t=0[/latex] and moving downward. Find a formula for the sinusoidal function [latex]h(t)[/latex] that gives the lever’s height.
- Find the first two times that lever is 1 meter above its lowest position.
Solution
- [latex]\displaystyle h(t)=-0.9\sin (20\pi t)+0.9[/latex]
- 0.0518 minutes and 0.9823 minutes
Concepts
-
Solutions of Trigonometric Equations.
- The equation [latex]\cos \theta = k, ~~ -1\lt k \lt 1{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]\theta_{1}=\cos^{-1}(k)~~ {and}~~ \theta_{2}=2\pi -\theta_{1}[/latex]
- The equation [latex]\sin \theta = k, ~~ -1\lt k \lt 1{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]{If}~k \gt 0:~~ \theta_{1} =\sin^{-1}(k)~~ {and}~~ \theta_{2}=\pi -\theta_{1}[/latex]
[latex]{If}~k \lt 0:~~ \theta_{1}=\sin^{-1}(k) + 2\pi ~~ {and}~~ \theta_{2}=\pi -\sin^{-1}(k)[/latex]
- The equation [latex]\tan \theta = k{,}[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi{:}[/latex]
[latex]{If}~k \gt 0:~~ \theta_{1}=\tan^{-1}(k)~~ {and}~~ \theta_{2}=\pi +\theta_{1}[/latex]
[latex]{If}~k \lt 0:~~ \theta_{1}=\tan^{-1}(k) + \pi ~~ {and}~~ \theta_{2}=\pi +\theta_{1}[/latex]
- If [latex]n[/latex] is a positive integer, the equations [latex]\sin n\theta = k[/latex] and [latex]\cos n\theta = k[/latex] each have [latex]2n[/latex] solutions between [latex]0[/latex] and [latex]2\pi{,}[/latex] for [latex]-1\lt k \lt 1{.}[/latex]
- The equation [latex]\tan n\theta = k[/latex] has one solution in each cycle of the graph.
- For more complicated equations, it can be helpful to use a substitution to replace the input of the trig function by a single variable.
-
Using a Substitution to Solve Trigonometric Equations.
To solve the equation [latex]\sin (Bx+C)=k[/latex] or [latex]\cos (Bx+C)=k{:}[/latex]
- Substitute [latex]\theta = Bx+C{,}[/latex] and find two solutions for [latex]\sin \theta = k[/latex] or [latex]\cos \theta = k{.}[/latex]
- Replace [latex]\theta[/latex] by [latex]Bx+C[/latex] in each solution, and solve for [latex]x{.}[/latex]
- Find the other solutions by adding multiples of [latex]\dfrac{2\pi}{B}[/latex] to the first two solutions.
To solve the equation [latex]\tan (Bx+C)=k{:}[/latex]
- Substitute [latex]\theta = Bx+C{,}[/latex] and find one solution for [latex]\tan \theta = k{.}[/latex]
- Replace [latex]\theta[/latex] by [latex]Bx+C[/latex] and solve for [latex]x{.}[/latex]
- Find the other solutions by adding multiples of [latex]\dfrac{\pi}{B}[/latex] to the first solution.
Study Questions
- If [latex]x=0.68[/latex] is one solution of the equation [latex]\cos x = c{,}[/latex] what is other? Illustrate on a unit circle.
- If [latex]t=2.45[/latex] is one solution of the equation [latex]\sin t = k{,}[/latex] what is the other? Illustrate on a unit circle.
- If [latex]\theta = 1.73[/latex] is one solution of the equation [latex]\tan \theta = C{,}[/latex] what is the other?
- Explain why the equation [latex]\cos nx = k,~~0 \lt k \lt 1{,}[/latex] has [latex]2n[/latex] solutions between [latex]0[/latex] and [latex]2\pi{.}[/latex]