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Solving Trig Problems: General Solutions
\(\displaystyle 6\cos \theta =3\sqrt{3}\)
\(\displaystyle \begin{array}{c}\cos \theta
=\frac{{\sqrt{3}}}{2}\\\\\left\{ {\theta |\theta
=\frac{\pi }{6}+2\pi k,\,\,\theta =\frac{{11\pi }}
{6}+2\pi k} \right\}\end{array}\)
\(\displaystyle \begin{array}{c}\color{#800000}{{\sin
x+3=4\sin x}}\\\\3\sin x=3\\\sin x=1\\\,\\\left\{ {x|x=\frac{\pi }
{2}+\,\,2\pi k} \right\}\end{array}\)
\(\displaystyle \begin{array}{c}\color{#800000}
{{\sqrt{3}\csc x=-2}}\\\text{csc}\,x=-\frac{2}
{{\sqrt{3}}}\,\,\,\,\,\,\,\,\,\left( {\sin x=-\frac{{\sqrt{3}}}
{2}} \right)\\\\\left\{ {x|x=\frac{{4\pi }}{3}+2\pi
k,\,x=\frac{{5\pi }}{3}+2\pi k} \right\}\end{array}\)
\(\displaystyle 2\sec \left( {x-\frac{\pi }{4}}
\right)=4\)
\(\displaystyle \begin{array}{c}\sec \left( {x-
\frac{\pi }{4}} \right)=2\text{ }\left( {\cos \left( {x-
\frac{\pi }{4}} \right)=\frac{1}{2}} \right)\\\\x-
\frac{\pi }{4}=\frac{\pi }{3}+2\pi k;\,\,x-\frac{\pi }
{4}=\frac{{5\pi }}{3}+2\pi k\\\left\{ {x|x=\frac{{7\pi
}}{{12}}+2\pi k,\,\,x=\frac{{23\pi }}{{12}}+2\pi k}
\right\}\end{array}\)
\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}
{c}4\sec \theta +10=-\sec \theta
\\\text{(degrees)}\end{array}}\\\\5\sec \theta =-10\\\sec \theta
=-2\text{ }\left( {\cos \theta =-\frac{1}{2}} \right)\\\\\left\{
\begin{array}{l}\theta |\theta =120{}^\circ +360{}^\circ
k,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,240{}^\circ +360{}^\circ
k\end{array} \right\}\end{array}\)
\(\displaystyle \begin{array}{c}\color{#800000}
{{2\sin x=4\sin x}}\\\\2\sin x=0\\\sin x=0\\\\\
{x|x=2\pi k,\,\,\,x=\pi +2\pi k\}\end{array}\)
Note that (by looking at Unit Circle) this can be
simplified to
\(\displaystyle \{x|x=\pi k\}\)
\(\displaystyle \text{co}{{\text{s}}^{2}}\theta
=\frac{1}{2}\)
\(\displaystyle \begin{array}{c}\sqrt{{\text{co}
{{\text{s}}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos
\theta =\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm
\frac{{\sqrt{2}}}{2}\\\\\{\theta |\theta =\frac{\pi }
{4}+2\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+2\pi
k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{5\pi }}{4}+2\pi
\(\displaystyle \tan \left( {x+\frac{\pi }{2}} \right)=\sqrt{3}\)
\(\displaystyle \begin{array}{c}x+\frac{\pi }{2}=\frac{\pi }
{3}\,\,\,\,\,\,\,\,\,\,\,\,x+\frac{\pi }{2}=\frac{{4\pi }}{3}\\\,\,\,x=-
\frac{\pi }{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5\pi }}
{6}\end{array}\)
Note that (by looking at Unit Circle) this can be simplified to
\(\displaystyle \begin{array}{c}\color{#800000}
{{3{{{\cot }}^{2}}\theta =1}}\\\text{co}
{{\text{t}}^{2}}\theta =\frac{1}{3}\\\sqrt{{{{{\cot
}}^{2}}\theta }}=\sqrt{{\frac{1}{3}}}\\\cot \theta
=\,\,\pm \frac{1}{{\sqrt{3}}}\\\\\{\theta |\theta
=\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi
k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{4\pi }}{3}+\pi
k,\,\,\theta =\frac{{5\pi }}{3}+\pi k\}\end{array}\)
k,\,\,\theta =\frac{{7\pi }}{4}+2\pi k\}\end{array}\)
Note that (by looking at Unit Circle) this can be
simplified to
\(\{\theta |\theta =\frac{\pi }{4}+\pi k,\,\,\,\theta
=\frac{{3\pi }}{4}+\pi k\}\)
\(\displaystyle \{x|x=\frac{{5\pi }}{6}+\pi k\}\)
(Remember that you add \(\pi k\) instead of \(2\pi k\) for tan
and cot).
Note that (by looking at Unit Circle) this can be
simplified to
\(\{\theta |\theta =\frac{\pi }{3}+\pi k,\,\,\,\theta
=\frac{{2\pi }}{3}+\pi k\}\)
\(\displaystyle \frac{{\cot \theta }}{4}=\sin \theta \cos \theta \)
\(\displaystyle \begin{array}{c}\frac{{\cos \theta }}{{4\sin \theta }}=\sin \theta
\cos \theta \\\cos \theta =4{{\sin }^{2}}\theta \cos \theta \\4{{\sin }^{2}}\theta
\cos \theta -\cos \theta =0\\\cos \theta \left( {4{{{\sin }}^{2}}-1} \right)=0\\\cos
\theta \left( {2\sin -1} \right)\left( {2\sin \theta +1} \right)=0\\\cos \theta =0\text{
}\,\,\text{ }\,\,\sin \theta =\pm \frac{1}{2}\,\\\\\{\theta |\theta =\frac{\pi }{2}+\pi
k,\,\,\,\theta =\frac{\pi }{6}+\pi k,\,\,\,\theta =\frac{{5\pi }}{6}+\pi k\,\}\end{array}\)
Watch domain restriction:
\(\displaystyle \frac{{1-\cos x}}{{\sin x}}=\frac{{\sin x}}{{1+\cos x}}\)
\(\begin{array}{c}\left( {1-\cos x} \right)\left( {1+\cos x} \right)={{\sin }^{2}}x\\1-
{{\cos }^{2}}x={{\sin }^{2}}x\\{{\sin }^{2}}x={{\sin
}^{2}}x;\,\,\,\,\mathbb{R}\end{array}\)
We get all real numbers since this is an
identity
. But we still need to check the
domain restrictions (denominator can’t be 0):
\(\begin{array}{c}\sin x\ne 0,\,\,\,1+\cos x\ne 0\\\sin x\ne 0,\,\,\,\cos x\ne -
1\\\text{Solution: }\left\{ {x|\,\,x\ne \pi +2\pi k} \right\}\end{array}\)
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