Solving Trigonometric Equations She Loves Math

Solving Trig Problems with Multiple Angles – General Solutions

\(\displaystyle \tan \left( {3\theta }

\right)=\sqrt{3}\)

 

\(3\theta =\frac{\pi }{3}+\pi

k\,\,\,\,\,\,\,\,3\theta =\frac{{4\pi }}{3}+\pi k\)

 

Looking at Unit Circle, this can be

simplified to:

\(\displaystyle \begin{array}{c}3\theta

=\frac{\pi }{3}+\pi k\\\left\{ {\theta |\theta

=\frac{\pi }{9}+\frac{\pi }{3}k}

\right\}\end{array}\)

\(\displaystyle 2\cos \left( {3x}

\right)+\sqrt{3}=0\)

 

\(\displaystyle \begin{array}{c}\cos \left(

{3x} \right)=-\frac{{\sqrt{3}}}

{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi

k\,\,\,\,\,\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi

k\\\\\left\{ {x|x=\frac{{5\pi }}{{18}}+\frac{2}

{3}\pi k,\,\,x=\frac{{7\pi }}{{18}}+\frac{2}

{3}\pi k\,\,} \right\}\end{array}\)

\(\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)-

2=0\)

 

\(\displaystyle \begin{array}{c}\sec \left( {\frac{x}{6}}

\right)=\frac{2}{{\sqrt{2}}}\,\,\,\,\,\,\,\,\left( {\cos \left(

{\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}}

\right)\\\,\,\frac{x}{6}=\frac{\pi }{4}+2\pi

k\,\,\,\,\,\,\,\,\,\,\,\,\frac{x}{6}=\frac{{7\pi }}{4}+2\pi

k\\\,\,\,x=\frac{{6\pi }}{4}+12\pi

k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{42\pi }}{4}+12\pi k\\\\\left\{

{x|x=\frac{{3\pi }}{2}+12\pi k,\,\,\,\,x=\frac{{21\pi }}

{2}+12\pi k\,\,} \right\}\end{array}\)

\(\displaystyle 2{{\sin }^{2}}\left( {2x}

---

\(\displaystyle 5{{\cos }^{2}}\left(

{\frac{\theta }{3}} \right)=5\)

 

\(\displaystyle \begin{array}{c}\sqrt{{{{{\cos

}}^{2}}\left( {\frac{\theta }{3}}

\right)}}=\sqrt{1}\\cos\left( {\frac{\theta }{3}}

\right)=\pm 1\\\,\,\,\frac{\theta }{3}=0+2\pi

k\,\,\,\,\,\,\,\,\frac{\theta }{3}=\pi +2\pi

k\end{array}\)

 

Looking at Unit Circle, this can be

simplified to:

\(\displaystyle \begin{array}{c}\frac{\theta }

{3}=\pi k\\\left\{ {\theta |\theta =3\pi k}

\right\}\end{array}\)

\right)=1\)

 

\(\displaystyle \begin{array}{c}{{\sin

}^{2}}\left( {2x} \right)=\frac{1}

{2}\\\sqrt{{{{{\sin }}^{2}}\left( {2x}

\right)}}=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x}

\right)=\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm

\frac{{\sqrt{2}}}{2}\\\,\,\,2x=\frac{\pi }

{4}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}

{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}{4}+2\pi

k\,\,\,\,\,\,\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi

k\end{array}\)

 

Looking at Unit Circle, this can be

simplified to:

\(\displaystyle \begin{array}{l}2x=\frac{\pi }

{4}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}

{4}+\pi k\\\left\{ {x|x=\frac{\pi }{8}+\frac{\pi }

{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k}

\right\}\end{array}\)

\(\displaystyle \tan \left( {\frac{\theta }{2}-\frac{\pi }{3}}

\right)=-1\)

 

\(\frac{\theta }{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi

k\,\,\,\,\,\,\,\,\,\frac{\theta }{2}-\frac{\pi }{3}=\frac{{7\pi

}}{4}+\pi k\)

 

Looking at Unit Circle, this can be simplified to:

\(\displaystyle \begin{array}{c}\frac{\theta }{2}-

\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\\\\\frac{\theta }

{2}=\left( {\frac{{3\pi }}{4}+\frac{\pi }{3}} \right)+\pi

k\\\frac{\theta }{2}=\frac{{13\pi }}{{12}}+\pi k\\\\\left\{

{\theta |\theta =\frac{{13\pi }}{6}+2\pi k}

\right\}\,\text{or}\,\left\{ {\theta |\theta =\frac{\pi }

{6}+2\pi k} \right\}\end{array}\)