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Factoring to Solve Trig Equations
\(\displaystyle \begin{array}{c}4\sin x\cos
x=2\sin x\\\text{Interval }(0,2\pi
)\end{array}\)
\(\displaystyle \begin{array}{c}4\sin x\cos
x-2\sin x=0\\2\sin x\left( {2\cos x-1}
\right)=0\\2\sin x=0\,\,\,\,\,\,\,\,\,\,2\cos x-
1=0\\\sin x=0\,\,\,\,\,\,\,\,\,\,\cos x=\frac{1}
{2}\\x=0,\,\,\pi \,\,\,\,\,\,\,\,x=\frac{\pi }
{3},\,\frac{{5\pi }}{3}\\\\\,x=\left\{ {0,\pi
,\frac{\pi }{3},\frac{{5\pi }}{3}}
\right\}\end{array}\)
\(\displaystyle \begin{array}{c}4{{\cos }^{4}}\theta
-7{{\cos }^{2}}\theta +3=0\\\text{General Solutions
(in degrees)}\end{array}\)
\(\displaystyle \begin{array}{c}\left( {4{{{\cos
}}^{2}}\theta -3} \right)\left( {{{{\cos }}^{2}}\theta -1}
\right)=0\\\,\sqrt{{{{{\cos }}^{2}}\theta
}}=\sqrt{{\frac{3}{4}}}\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\cos
}}^{2}}\theta }}=\sqrt{1}\\\,\cos \theta =\pm
\frac{{\sqrt{3}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta
=\pm 1\,\\\text{(Add }180{}^\circ k\text{ instead of
}360{}^\circ k\\\text{ because of }\pm )\\\,\,\,\theta
=30{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta
=180{}^\circ k\\\,\theta =330{}^\circ +180{}^\circ
k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left\{
\begin{array}{l}\theta |\theta =30{}^\circ
+180{}^\circ k,\,\,\theta =330{}^\circ +180{}^\circ
k,\\\,\,\,\,\theta =180{}^\circ k\end{array}
\right\}\end{array}\)
\(\displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec
x=2\\\text{General Solutions}\end{array}\)
\(\displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec x-
2=0\\\left( {2\sec x+1} \right)\left( {\sec x-2}
\right)=0\\\,\,\,\sec x=-\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec
x=2\\\text{(no solution)}\,\,\,\,\,\,\,\,\,\,\,\,\left( {\cos x=\frac{1}
{2}} \right)\\\\\left\{ {x|x=\frac{\pi }{3}+2\pi k,x=\frac{{5\pi }}
{3}+2\pi k\,} \right\}\end{array}\)
\(\displaystyle \begin{array}{c}{{\tan
}^{4}}x-4{{\tan }^{2}}x+3=0\\\text{General
Solutions}\end{array}\)
\(\displaystyle \begin{array}{c}\left( {{{{\tan
}}^{2}}x-3} \right)\left( {{{{\tan }}^{2}}x-1}
\right)=0\\{{\tan }^{2}}x-3=0\,\,\,\,\,\,\,{{\tan
\(\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot
x-\sqrt{3}\cot x=\sqrt{3}\\\text{Interval }(0,2\pi
)\end{array}\)
\(\displaystyle \begin{array}{c}\sqrt{3}\sin \left( {2\theta }
\right)\cot \left( {2\theta } \right)-\sin \left( {2\theta }
\right)=0\\\text{Interval }(0,2\pi )\end{array}\)
\(\displaystyle \begin{array}{c}\sin \left( {2\theta }
\right)\left( {\sqrt{3}\cot \left( {2\theta } \right)-1}
\right)=0\\\sin \left( {2\theta } \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\cot
\left( {2\theta } \right)=\frac{1}{{\sqrt{3}}}\\2\theta =\pi
}^{2}}x-1=0\\\sqrt{{{{{\tan
}}^{2}}x}}=\sqrt{3}\,\,\,\,\,\,\,\sqrt{{{{{\tan
}}^{2}}x}}=\sqrt{1}\\\tan x=\pm
\sqrt{3}\,\,\,\,\,\,\,\,\tan x=\pm
\sqrt{1}\\x=\frac{\pi }{3}+\pi
k\,\,\,\,\,\,\,\,\,\,x=\frac{\pi }{4}+\pi
k\\x=\frac{{2\pi }}{3}+\pi
k\,\,\,\,\,\,\,\,\,\,x=\frac{{3\pi }}{4}+\pi
k\\\\\left\{ \begin{array}{l}x|x=\frac{\pi }
{3}+\pi k,\,\,\,x=\frac{{2\pi }}{3}+\pi
k,\\\,\,\,\,x=\frac{\pi }{4}+\pi
k,\,\,\,x=\frac{{3\pi }}{4}+\pi k\end{array}
\right\}\end{array}\)
\(\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot
x-\sqrt{3}\cot x-\sqrt{3}=0\\3\cot x\left( {\cot x+1}
\right)-\sqrt{3}\left( {\cot x+1} \right)=0\\\left( {\cot
x+1} \right)\left( {3\cot x-\sqrt{3}} \right)=0\\\,\,\cot
x=-1\,\,\,\,\,\,\,\,\,\,\,\cot x=\frac{{\sqrt{3}}}
{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan x=-1\,\,\,\,\,\,\,\,\,\tan
x=\frac{3}{{\sqrt{3}}}=\sqrt{3}\\\,\,\,x=\frac{{3\pi }}
{4},\,\,\frac{{7\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,x=\frac{\pi }
{3},\,\,\frac{{4\pi }}{3}\\\\x=\left\{ {\frac{{3\pi }}
{4},\frac{{7\pi }}{4},\frac{\pi }{3},\frac{{4\pi }}{3}}
\right\}\end{array}\)
k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\theta =\frac{\pi }{3}+\pi
k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\theta
=\frac{{4\pi }}{3}+\pi k\\\,\,\,\theta =\frac{{\pi k}}
{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{\pi }{6}+\frac{\pi }
{2}k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta
=\frac{{4\pi }}{6}+\frac{\pi }{2}k\end{array}\)
Because of the domain restriction for cot (where its
asymptotes are), and noting that \(\cot \left[ {2\left(
{\frac{{\pi k}}{2}} \right)} \right]=\cot \left( {\pi k} \right)\) is
undefined, we have to eliminate \(\frac{{\pi k}}{2}\).
Solutions are:
\(\theta =\left\{ {\frac{\pi }{6},\frac{{2\pi }}{3},\frac{{7\pi }}
{6},\frac{{5\pi }}{3}} \right\}\)
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