# Solving Trigonometry Problems

Also note that  is written as $$\theta$$, and we can put it in the graphing calculator as $$\boldsymbol$$ or $$\boldsymbol$$., we have to first find the general solution of the equation, and then go back to the Unit Circle to see where the solutions are in the $$\left[ 0,2\pi \right)$$ interval.We will learn how to do this Note that sometimes you may have to solve using degrees $$\left[ \right)$$ instead of radians.

Also note that  is written as $$\theta$$, and we can put it in the graphing calculator as $$\boldsymbol$$ or $$\boldsymbol$$., we have to first find the general solution of the equation, and then go back to the Unit Circle to see where the solutions are in the $$\left[ 0,2\pi \right)$$ interval.We will learn how to do this Note that sometimes you may have to solve using degrees $$\left[ \right)$$ instead of radians.$$\displaystyle \begin\sec \left( \right)=\frac\,\,\,\,\,\,\,\,\left( \right)\\,\,\frac=\frac 2\pi k\,\,\,\,\,\,\,\,\,\,\,\,\frac=\frac 2\pi k\\,\,\,x=\frac 12\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac 12\pi k\\\left\end$$ $$\displaystyle \begin\sqrt=\sqrt\cos\left( \right)=\pm 1\\,\,\,\frac=0 2\pi k\,\,\,\,\,\,\,\,\frac=\pi 2\pi k\end$$ Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin\frac=\pi k\\left\end$$ $$\displaystyle \begin\left( \right)=\frac\\sqrt=\sqrt\\sin \left( \right)=\,\,\pm \frac=\,\,\pm \frac\\,\,\,2x=\frac 2\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac 2\pi k\\,\,\,2x=\frac 2\pi k\,\,\,\,\,\,\,\,\,\,2x=\frac 2\pi k\end$$ Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin2x=\frac \pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac \pi k\\left\end$$ $$\frac-\frac=\frac \pi k\,\,\,\,\,\,\,\,\,\frac-\frac=\frac \pi k$$ Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin\frac-\frac=\frac \pi k\\\frac=\left( \right) \pi k\\frac=\frac \pi k\\\left\,\text\,\left\end$$ $$3\theta =\frac \pi k\,\,\,\,\,\,\,\,3\theta =\frac \pi k$$ Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin3\theta =\frac \pi k\\,\theta =\frac \frack\,\,=\,\,\frac \frack\end$$ Looking at the Unit Circle, we can get solutions between $$\displaystyle \begin\cos \left( \right)=-\frac\\,\,\,3x=\frac 2\pi k\,\,\,\,\,\,\,\,\,3x=\frac 2\pi k\\,\,\,x=\frac \frac\pi k\,\,\,\,\,\,\,\,\,\,x=\frac \frac\pi k\end$$ Looking at the Unit Circle, we can get solutions between $$\displaystyle \begin\sec \left( \right)=\frac\,\,\,\,\,\,\,\left( \right)\\,\,\frac=\frac 2\pi k\,\,\,\,\,\,\,\,\,\,\,\frac=\frac 2\pi k\\,\,\,x=\frac 12\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac 12\pi k\end$$ Looking at the Unit Circle, the only solution that will work is $$\frac=\frac$$.

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We learned how to factor Quadratic Equations in the Solving Quadratics by Factoring and Completing the Square section.

Note that when we factor trig equations to find solutions, like we do with “regular” equations, we never just divide a factor out from each side.

Note that we will use Trigonometric Identities to solve trig problems in the Trigonometric Identity section.

Important Note: There is a subtle distinction between finding inverse trig functions and solving for trig functions.

Note that $$k$$ represents all integers $$\left( k\in \mathbb \right)$$.

Note also that I’m using “fancy” notation; you may not be required to do this., depending on your book or teacher.

In doing this, we are probably “throwing away” valid solutions to the equation.

Here are some examples, both solving on the interval and over the reals; note one of the problems is using degrees instead of radians.

This will sometimes happen if trig functions are squared in the problems also, since we’ll getting plusses and minuses.

Here are examples; find the general solution, or all real solutions) for the following equations.

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