8.7E Exercises
- Page ID
- 158839
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Give at least three specific solutions to the equation, and describe all possible solutions.
- \( - \sin \theta + 1 = 0 \)
- \( \sqrt{2} \cos x = 1 \)
- \( \tan \theta = -1 \)
- \( \sin \left( x + \frac{\pi}{2} \right) = \frac{1}{2} \)
- \( \cos^2 \theta = 1\)
- \( \cot (4x) = \frac{1}{\sqrt{3}} \) (Hint: what must \(\tan 4x\) be for this equation to be true?)
- Answer
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- \( \theta = -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, ...\) aka \( \theta = \frac{\pi}{2} + 2k\pi\) where \(k\) is any integer.
- \( x = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{7\pi}{4},...\) aka \( \theta = \frac{\pi}{4}+2k\pi, \theta = \frac{7\pi}{4} + 2k\pi \) where \(k\) is any integer.
- \( \theta = \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, ...\) aka \( \theta = \frac{3\pi}{4}+k\pi \) where \(k\) is any integer.
- \( \theta = -\frac{\pi}{3}, \frac{5\pi}{3}, ...\) and \( \theta = \frac{\pi}{3}, \frac{7\pi}{3}, ...\) aka \( \theta = \frac{\pi}{3}+2k\pi, \theta = \frac{5\pi}{3}+2k\pi\), where \(k\) is any integer.
- Hint: there are two ways that \( (\cos \theta)^2\) can turn out to be positive \(1\). It turns out \( \theta = 0, \pi, 2\pi, 3\pi, ...\) aka \( \theta = k\pi\), where \(k\) is any integer, or "multiples of \(\pi\)."
- \( 4x = \frac{\pi}{3}+\pi k \), so \( x = \frac{1}{4} \left( \frac{\pi}{3}+\pi k \right) \), of which some examples would be \( \frac{\pi}{12} \) (if \(k=0\)), \(\frac{\pi}{3}\) (if \(k = 1\)), and \(\frac{7\pi}{12}\) (if \(k = 2\)), etc.
Find all solutions.
- \( \sin^2 \theta - 1 = 0\)
- \( \cos^2 \theta + 2\cos \theta = -1 \)
- \( \tan \theta \sin \theta + \sin \theta = 0\)
- \( 2\sin \theta \cos \theta - \cos \theta = 0\)
- Answer
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- \(\theta = \frac{\pi}{2}+2k\pi \), \(\theta = \frac{3\pi}{2}+2k\pi\) where \(k\) is any integer. These can be combined as \(\theta = \frac{\pi}{2}+k\pi \). To convince yourself of this, write out a list of options for different choices of \(k\) and compare. (For example, if \(k = 0\), we get \( \frac{\pi}{2}, \frac{3\pi}{2}\)...if \(k = 1\), we get \( \frac{5\pi}{2}, \frac{7\pi}{2}\)... Keep going. You will get all "odd multiples" of \(\frac{\pi}{2}\), which are also covered by the formula \( \frac{\pi}{2} + k \pi\), if you write out a list of HIS examples.)
- \( \theta = \pi + 2k\pi\) where \(k\) is any integer
- \( \theta = k\pi\) and \( \theta = \frac{3\pi}{4}+k\pi\), where \(k\) is any integer.
- \( \theta = \frac{\pi}{2}+2k\pi, \theta = \frac{3\pi}{2}+2k\pi\) and \(\theta = \frac{\pi}{6}+2k\pi, \theta = \frac{5\pi}{6} + 2k\pi\) where \(k\) is any integer.
Give at least three specific solutions to the equation, and describe all possible solutions.
- \(2 \sin \theta \cos \theta = 0 \)
- \( \dfrac{\sin^2 \theta}{\cos \theta} + \cos \theta = 1 \)
- \( \tan x = \sec^2 x - 1\)
- Answer
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- Hint: this is one half of a certain identity, so it's equivalent to something simpler... Turns out to be \(\theta = k\frac{\pi}{2}\) where \(k\) is any integer.
- Hint: common denominator... \( \theta = 2k\pi \) where \(k\) is any integer.
- \( \theta = \frac{\pi}{4}+k\pi\), \(\theta = k\pi\) where \(k\) is any integer.
Follow the steps to find all solutions to the equation
\[ \left| \tan \theta - \frac{2}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}} \notag \]
- Recall that the effect of absolute value bars is split into two cases:
\[ |\text{inside}| = \begin{cases} \text{inside} & \text{ if inside }\geq 0 \\ -(\text{inside}) & \text{ if inside }<0 \end{cases} \notag \]
Use this fact to split this equation into two cases. - Find the solutions to both case equations with the usual techniques.
- Answer
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- The cases are \( \tan \theta - \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \) and \( -(\tan \theta- \frac{2}{\sqrt{3}}) = \frac{1}{\sqrt{3}}\). Simplify these equations so that you can solve them.
- Hint: you can write \( \frac{3}{\sqrt{3}}\) as \( \sqrt{3}\). \( \theta = \frac{\pi}{6} + k\pi, \theta = \frac{\pi}{3} + k\pi,\) where \(k\) is any integer.