Solving Trigonometric Equations She Loves Math

Solving Trig Problems:   \(\boldsymbol {\left[ {0,2\pi } \right)}\)

\(\displaystyle \sin \theta =1\)

\(\theta =\left\{ {\frac{\pi }{2}}

\right\}\)

 

\(\displaystyle \sin \theta =0\)

\(\theta =\left\{ {0,\pi }

\right\}\)

 

\(\displaystyle \cos \theta =0\)

\(\theta =\left\{ {\frac{\pi }

{2},\frac{{3\pi }}{2}} \right\}\)

\(\tan \theta +1=0\)

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

=-1\\\theta =\left\{ {\frac{{3\pi }}

{4},\frac{{7\pi }}{4}} \right\}\end{array}\)

 

\(\tan \theta -1=0\)

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

=1\\\theta =\left\{ {\frac{\pi }{4},\frac{{5\pi

}}{4}} \right\}\end{array}\)

\(\displaystyle \begin{array}{c}3\sqrt{2}\cos \theta

+3=0\\\text{(degrees)}\end{array}\)

 

\(\displaystyle \begin{array}{c}3\sqrt{2}\cos \theta =-3\\\cos

\theta =-\frac{3}{{3\sqrt{2}}}\\\cos \theta =-\frac{1}{{\sqrt{2}}}=-

\frac{{\sqrt{2}}}{2}\\\theta =\left\{ {135{}^\circ ,225{}^\circ }

\right\}\end{array}\)

\(2\sin \theta +4=3\)

 

\(\begin{array}{c}2\sin \theta =-

1\\\sin \theta =-\frac{1}{2}\\\theta

=\left\{ {\frac{{7\pi }}{6},\frac{{11\pi

}}{6}} \right\}\end{array}\)

\(\displaystyle 4\sec \theta

+7=-1\)

 

\(\displaystyle \begin{array}

{c}4\sec \theta =-8\\\sec

\theta =-2\,\,\,\,\,\left( {\cos

\theta =-\frac{1}{2}}

\right)\\\theta =\left\{

\(\displaystyle 2{{\cos }^{2}}\theta =1\)

 

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

}^{2}}\theta =\frac{1}{2}\\\sqrt{{{{{\cos

}}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos

\theta =\pm \frac{1}{{\sqrt{2}}}=\pm

\frac{{\sqrt{2}}}{2}\\\theta =\left\{ {\frac{\pi

}{4},\frac{{3\pi }}{4},\frac{{5\pi }}

\(\displaystyle \sin \left( {\theta +\frac{\pi }{{18}}} \right)=0\)

 

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

{{18}}=0\,\,\,\,\,\,\theta +\frac{\pi }{{18}}=\pi \\\,\,\,\theta =-\frac{\pi

}{{18}}\,\,\,\,\,\,\,\theta =\frac{{17\pi }}{{18}}\\\text{Since}-\frac{\pi

}{{18}}\text{ isn }\!\!’\!\!\text{ t in}\left[{0,\text{2}\pi } \right),\\\left\{

{-\frac{\pi }{{18}}+2\pi ,\frac{{17\pi }}{{18}}} \right\}\\=\left\{

\(\displaystyle \sin \left( {\frac{\theta

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

{\frac{\theta }{3}+\frac{\pi }{3}}

\right)\)

 

\(\displaystyle \begin{array}{c}\tan

\left( {\frac{\theta }{3}+\frac{\pi }{3}}

\right)=1\\\frac{\theta }{3}+\frac{\pi }

{3}=\frac{\pi }

{4}\,\,\,\,\,\,\,\,\,\,\frac{\theta }

{3}+\frac{\pi }{3}=\frac{{5\pi }}

{4}\\\,\,\,\frac{\theta }{3}=-\frac{\pi }

{{12}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\theta

}{3}=\frac{{11\pi }}{{12}}\\\,\,\,\theta

=-\frac{\pi }

---

{\frac{{2\pi }}{3},\frac{{4\pi }}

{3}} \right\}\end{array}\)

{4},\frac{{7\pi }}{4}} \right\}\end{array}\)

{\frac{{17\pi }}{{18}},\frac{{35\pi }}{{18}}} \right\}\end{array}\)

{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta

=\frac{{33\pi }}{{12}}\end{array}\)

 

Neither are in \(\left[ {0,2\pi }

\right)\), and adding \(2\pi \) to \

(\displaystyle -\frac{\pi }{4}\)

doesn’t work (extraneous); no

solution, or \(\emptyset\).

Here’s one where we have an  inequality:

 

For what values of \(\theta \) is \(2\cos \theta <-\sqrt{3}\) on the \(\left[

{0,2\pi } \right)\) interval?

 

\(\displaystyle 2\cos \theta <-\sqrt{3}\)

\(\displaystyle \cos \theta <-\frac{{\sqrt{3}}}{2}\)

This is a little tricky; if we look on the Unit Circle, we want the  cos values to be less than where \

(\displaystyle \cos \theta =-\frac{{\sqrt{3}}}{2}\). Since when \(\displaystyle \cos \theta =-

\frac{{\sqrt{3}}}{2},\,\,\theta =\left\{ {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right\}\), the \(x\) will be less (or

to the left of) of those 2 values, so we can see that \(\displaystyle \frac{{5\pi }}{6}<\theta <\frac{{7\pi

}}{6}\text{ or }\left( {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right)\).

 

You can also plug in values for  cos on the Unit Circle to see which are less than \(\displaystyle -

\frac{{\sqrt{3}}}{2}\), or use a Graphing Calculator to graph the two sides of the inequality.