Solving a trigonometric equation is just like solving a regular equation. You will use factoring and other algebraic techniques to get the variable on one side. The biggest difference with trigonometric equations is the opportunity for there to be an infinite number of solutions that must be described with a pattern. The equation \begin{align*}\cos x=1\end{align*} has many solutions including 0 and \begin{align*}2 \pi\end{align*}. How would you describe all of them?

http://www.youtube.com/watch?v=ABKO3ta_Azw James Sousa: Solve Trigonometric Equations II

http://www.youtube.com/watch?v=7thuFLqC7z0 James Sousa: Solve Trigonometric Equations III

#### Guidance

The identities you have learned are helpful in solving trigonometric equations. The goal of solving an equation hasn’t changed. Do whatever it takes to get the variable alone on one side of the equation. Factoring, especially with the Pythagorean identity, is critical.

When solving trigonometric equations, try to give exact (non-rounded) answers. If you are working with a calculator, keep in mind that while some newer calculators can provide exact answers like \begin{align*}\frac{\sqrt{3}}{2}\end{align*}, most calculators will produce a decimal of 0.866... If you see a decimal like 0.866..., try squaring it. The result might be a nice fraction like \begin{align*}\frac{3}{4}\end{align*}. Then you can logically conclude that the original decimal must be the square root of \begin{align*}\frac{3}{4}\end{align*} or \begin{align*}\frac{\sqrt{3}}{2}\end{align*}.

When solving, if the two sides of the equation are always equal, then the equation is an identity. If the two sides of an equation are never equal, as with \begin{align*}\sin x=3\end{align*}, then the equation has no solution.

**Example A**

Solve the following equation algebraically and confirm graphically on the interval \begin{align*}[-2 \pi, 2 \pi]\end{align*}.

\begin{align*}\cos 2x=\sin x\end{align*}

**Solution:**

**\begin{align*}\cos 2x &= \sin x\\
1-2 \sin^2 x &= \sin x\\
0 &= 2 \sin^2 x+\sin x-1\\
0 &= (2 \sin x-1)(\sin x+1)\end{align*}**

Solving the first part set equal to zero within the interval yields:

\begin{align*}0 &= 2 \sin x-1\\ \frac{1}{2} &= \sin x\\ x &= \frac{\pi}{6}, \frac{5 \pi}{6}, - \frac{11 \pi}{6}, -\frac{7 \pi}{6}\end{align*}

Solving the second part set equal to zero yields:

\begin{align*}0 &= \sin x+1\\ -1 &= \sin x\\ x &= -\frac{\pi}{2}, \frac{3 \pi}{2}\end{align*}

These are the six solutions that will appear as intersections of the two graphs \begin{align*}f(x)=\cos 2x\end{align*} and \begin{align*}g(x)=\sin x\end{align*}.

**Example B**

Determine the general solution to the following equation.

\begin{align*}\cot x-1=0\end{align*}

**Solution: **

**\begin{align*}\cot x-1 &= 0\\
\cot x &= 1\end{align*}**

One solution is \begin{align*}x=\frac{\pi}{4}\end{align*}. However, since this question asks for the general solution, you need to find every possible solution. You have to know that cotangent has a period of \begin{align*}\pi\end{align*} which means if you add or subtract \begin{align*}\pi\end{align*} from \begin{align*}\frac{\pi}{4}\end{align*} then it will also yield a height of 1. To capture all these other possible \begin{align*}x\end{align*} values you should use this notation.

\begin{align*}x=\frac{\pi}{4} \pm n \cdot \pi\end{align*} where \begin{align*}n\end{align*} is a integer

**Example C**

Solve the following equation.

\begin{align*}4 \cos^2 x-1=3-4 \sin^2 x\end{align*}

**Solution:**

**\begin{align*}4 \cos^2 x-1 &= 3-4 \sin^2 x\\
4 \cos^2 x+4 \sin^2 x &= 3+1\\
4 (\cos^2 x+\sin^2 x) &= 4\\
4 &= 4\end{align*}**

This equation is always true which means the right side is always equal to the left side. This is an identity.

**Concept Problem Revisited**

The equation \begin{align*}\cos x=1\end{align*} has many solutions. When you type \begin{align*}\cos^{-1} 1\end{align*} on your calculator, it will yield only one solution which is 0. In order to describe all the solutions you must use logic and the graph to figure out that cosine also has a height of 1 at \begin{align*}-2 \pi, 2 \pi, -4 \pi, 4 \pi \ldots\end{align*} Luckily all these values are sequences in a clear pattern so you can describe them all in general with the following notation:

\begin{align*}x=0 \pm n \cdot 2 \pi\end{align*} where \begin{align*}n\end{align*} is an integer, or \begin{align*}x=\pm n \cdot 2 \pi\end{align*} where \begin{align*}n\end{align*} is an integer

#### Vocabulary

The terms ** “general solution,” “completely solve”**, and

**mean you must find solutions to an equation without the use of a calculator. In addition, trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions. When you see these directions remember to find all the solutions by using notation like in Example B.**

*“solve exactly”*#### Guided Practice

1. Solve the following equation on the interval \begin{align*}(2 \pi, 4 \pi)\end{align*}.

\begin{align*}2 \sin x+1=0\end{align*}

2. Solve the following equation exactly.

\begin{align*}2 \cos^2 x+3 \cos x-2=0\end{align*}

3. Create an equation that has the solutions:

\begin{align*}\frac{\pi}{4} \pm n \cdot 2 \pi\end{align*} where \begin{align*}n\end{align*} is an integer

**Answers:**

1. First, solve for the solutions within one period and then use logic to find the solutions in the correct interval.

\begin{align*}2 \sin x+1 &= 0\\ \sin x &= -\frac{1}{2}\\ x &= \frac{7 \pi}{6}, \frac{11 \pi}{6}\end{align*}

You must add \begin{align*}2 \pi\end{align*} to each of these solutions to get solutions that are in the interval.

\begin{align*}x=\frac{19 \pi}{6}, \frac{23 \pi}{6}\end{align*}

2. Start by factoring:

\begin{align*}2 \cos^2 x+3 \cos x-2 &= 0\\ (2 \cos x-1)(\cos x+2) &= 0\end{align*}

Note that \begin{align*}\cos x \neq -2\end{align*} which means only one equation needs to be solved for solutions.

\begin{align*}2 \cos x-1 &= 0\\ \cos x &= \frac{1}{2}\\ x&=\frac{\pi}{3}, -\frac{\pi}{3}\end{align*}

These are the solutions within the interval \begin{align*}-\pi\end{align*} to \begin{align*}\pi\end{align*}. Since this represents one full period of cosine, the rest of the solutions are just multiples of \begin{align*}2 \pi\end{align*} added and subtracted to these two values.

\begin{align*}x=\pm \frac{\pi}{3} \pm n \cdot 2 \pi\end{align*} where \begin{align*}n\end{align*} is an integer

3. There are an infinite number of possible equations that will work. When you see the \begin{align*}\frac{\pi}{4}\end{align*} you should think either of where tangent is equal to one or where sine/cosine is equal to \begin{align*}\frac{\sqrt{2}}{2}\end{align*}. The problem with both of these initial guesses is that tangent repeats every \begin{align*}\pi\end{align*} not every \begin{align*}2 \pi\end{align*}, and sine/cosine have a second place where they reach a height of 1. An option that works is:

\begin{align*}\tan \frac{x}{2}=1\end{align*}

This equation works because the period of \begin{align*}\tan \frac{x}{2}\end{align*} is \begin{align*}2 \pi\end{align*}.

#### Practice

Solve each equation on the interval \begin{align*}[0, 2 \pi)\end{align*}.

1. \begin{align*}3 \cos^2 \frac{x}{2}=3\end{align*}

2. \begin{align*}4 \sin^2 x=8 \sin^2 \frac{x}{2}\end{align*}

Find approximate solutions to each equation on the interval \begin{align*}[0, 2 \pi)\end{align*}.

3. \begin{align*}3 \cos^2 x+10 \cos x+2=0\end{align*}

4. \begin{align*}\sin^2 x+3 \sin x=5\end{align*}

5. \begin{align*}\tan^2 x+\tan x=3\end{align*}

6. \begin{align*}\cot^2 x+5 \tan x+14=0\end{align*}

7. \begin{align*}\sin^2 x+\cos^2 x=1\end{align*}

Solve each equation on the interval \begin{align*}[0,360^\circ)\end{align*}.

8. \begin{align*}2 \sin \left(x-\frac{\pi}{2}\right)=1\end{align*}

9. \begin{align*}4 \cos (x-\pi)=4\end{align*}

Solve each equation on the interval \begin{align*}[2 \pi, 4 \pi)\end{align*}.

10. \begin{align*}\cos^2 x+2 \cos x+1=0\end{align*}

11. \begin{align*}3 \sin x=2 \cos^2 x\end{align*}

12. \begin{align*}\tan x \sin^2 x=\tan x\end{align*}

13. \begin{align*}\sin^2 x+1=2 \sin x\end{align*}

14. \begin{align*}\sec^2 x=4\end{align*}

15. \begin{align*}\sin^2 x-4=\cos^2 x-\cos 2x-4\end{align*}