Skip Navigation

Trigonometric Equations Using Factoring

Factoring and the Quadratic Formula.

Atoms Practice
Estimated17 minsto complete
Practice Trigonometric Equations Using Factoring
This indicates how strong in your memory this concept is
Estimated17 minsto complete
Practice Now
Turn In
Solving Trigonometric Equations Using Quadratic Techniques

As Agent Trigonometry, you are given this clue: \begin{align*}2\cos^2 x-3 \cos x + 1=0\end{align*}. If \begin{align*}x\end{align*} falls in the interval \begin{align*}0 \le x < 2\pi\end{align*}, what is/are its possible value(s)?

Solving Trigonometric Equations 

Another way to solve a trig equation is to use factoring or the quadratic formula. Let’s look at a few problems.

Let's solve the following trigonometric equations.

  1. Solve \begin{align*}\sin^2x-3 \sin x+2=0\end{align*}.

This sine equation looks a lot like the quadratic \begin{align*}x^2-3x+2=0\end{align*} which factors to be \begin{align*}(x-2)(x-1)=0\end{align*} and the solutions are \begin{align*}x = 2\end{align*} and 1. We can factor the trig equation in the exact same manner. Instead of just \begin{align*}x\end{align*}, we will have \begin{align*}\sin x\end{align*} in the factors.

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

There is no solution for \begin{align*}\sin x=2\end{align*} and \begin{align*}\sin x=1\end{align*} when \begin{align*}x= \frac{\pi}{2} \pm 2 \pi n\end{align*}.

  1. Solve \begin{align*}1- \sin x=\sqrt{3} \cos x\end{align*} in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

To solve this equation, use the Pythagorean Identity \begin{align*}\sin^2x+\cos^2x=1\end{align*}. Solve for either cosine and substitute into the equation. \begin{align*}\cos x=\sqrt{1- \sin^2 x}\end{align*}

\begin{align*}1- \sin x&=\sqrt{3} \cdot \sqrt{1- \sin^2x} \\ (1- \sin x)^2&=\sqrt{3-3 \sin^2x}^2 \\ 1-2 \sin x+\sin^2x&=3-3 \sin^2x \\ 4 \sin^2x-2 \sin x-2&=0 \\ 2 \sin^2x-\sin x-1&=0 \\ (2 \sin x+1)(\sin x-1)&=0\end{align*}

Solving each factor for \begin{align*}x\end{align*}, we get \begin{align*}\sin x=- \frac{1}{2} \rightarrow x=\frac{7 \pi}{6}\end{align*} and \begin{align*}\frac{11 \pi}{6}\end{align*} and \begin{align*}\sin x=1 \rightarrow x=\frac{\pi}{2}\end{align*}.

  1. Solve \begin{align*}\tan^2x-5 \tan x-9=0\end{align*} in the interval \begin{align*}0 \le x < \pi\end{align*}.

This equation is not factorable so you have to use the Quadratic Formula.

\begin{align*}\tan x&=\frac{5 \pm \sqrt{\left(-5\right)^2-4 \left(1\right) \left(-9\right)}}{2} \\ &=\frac{5 \pm \sqrt{61}}{2} \\ & \approx 6.41 \ and \ -1.41\end{align*}

\begin{align*}x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}\end{align*} and \begin{align*}x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}\end{align*}

The first answer is within the range, but the second is not. To adjust -0.954 to be within the range, we need to find the answer in the second quadrant, \begin{align*}\pi - 0.954=2.186 \ \text{rad}\end{align*}.


Example 1

Earlier, you were asked to find the possible values for \begin{align*}2\cos^2 x-3 \cos x + 1=0\end{align*}.

We can solve this problem by factoring.

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

Over the interval \begin{align*}0 \le x < 2\pi\end{align*}, \begin{align*}\cos x = \frac{1}{2}\end{align*} when \begin{align*}x = \frac{\pi}{3}\end{align*} and \begin{align*}\cos x = 1\end{align*} when \begin{align*}x = 0\end{align*}.

Solve the following trig equations using any method in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

Example 2

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

Put everything onto one side of the equation and factor out a cosine.

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

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

Example 3

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

Recall that \begin{align*}\sin(-x)=- \sin x\end{align*} from the Negative Angle Identities.

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

Example 4

\begin{align*}4 \cos^2x-2 \cos x-1=0\end{align*}

This quadratic is not factorable, so use the quadratic formula.

\begin{align*}\cos x&=\frac{2 \pm \sqrt{2^2 -4 \left(4\right) \left(-1\right)}}{2 \left(4\right)} \\ &=\frac{2 \pm \sqrt{20}}{8} \\ &=\frac{1 \pm 2 \sqrt{5}}{4}\end{align*}

\begin{align*}x& \approx \cos^{-1} \left(\frac{1+ \sqrt{5}}{4}\right) && x \approx \cos^{-1} \left(\frac{1- \sqrt{5}}{4}\right) \\ & \approx \cos^{-1} 0.8090 \qquad and && \ \ \approx \cos^{-1}-0.3090 \\ & \approx 0.6283 && \ \ \approx 1.8850 \ (\text{reference angle is} \ \pi-1.8850 \approx 1.2570)\end{align*}

The other solutions in the range are \begin{align*}x \approx 2 \pi - 0.6283 \approx 5.6549\end{align*} and \begin{align*}x \approx \pi + 1.2570 \approx 4.3982\end{align*}.


Solve the following trig equations using any method. Find all solutions in the interval \begin{align*}0 \le x < 2 \pi\end{align*}. Round any decimal answers to 4 decimal places.

  1. \begin{align*}2 \cos^2x-\sin x -1=0\end{align*}
  2. \begin{align*}4 \sin^2x+5 \sin x+1=0\end{align*}
  3. \begin{align*}3 \tan^2x- \tan x=0\end{align*}
  4. \begin{align*}2 \cos^2x+\cos(-x)-1=0\end{align*}
  5. \begin{align*}1- \sin x=\sqrt{2} \cos x\end{align*}
  6. \begin{align*}\sqrt{\sin x}=2 \sin x-1\end{align*}
  7. \begin{align*}\sin^3x-\sin x=0\end{align*}
  8. \begin{align*}\tan^2x-8 \tan x+7=0\end{align*}
  9. \begin{align*}5 \cos^2x+3 \cos x-2=0\end{align*}
  10. \begin{align*}\sin x- \sin x \cos^2x=1\end{align*}
  11. \begin{align*}\cos^2x-3 \cos x+2=0\end{align*}
  12. \begin{align*}\sin^2x \cos x=4 \cos x\end{align*}
  13. \begin{align*}\cos x \csc^2x+2 \cos x=6 \cos x\end{align*}

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

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

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.11. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Trigonometric Equations Using Factoring.
Please wait...
Please wait...