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Trigonometric Equations Using Factoring

Factoring and the Quadratic Formula.

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Practice Trigonometric Equations Using Factoring
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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)?


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

Example A

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

Solution: 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*}.

Example B

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

Solution: 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*}.

Example C

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

Solution: 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*}.

Concept Problem Revisit 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*}.

Guided Practice

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

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

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

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


1. 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*}

2. 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*}

3. 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*}.

Explore More

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*}

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