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3.5: Trigonometric Equations Using the Quadratic Formula

Difficulty Level: At Grade Created by: CK-12
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Solving equations is a fundamental part of mathematics. Being able to find which values of a variable fit an equation allows us to determine all sorts of interesting behavior, both in math and in the sciences. Solving trig equations for angles that satisfy the equation is one application of mathematical methods for solving equations. Suppose someone gave you the following equation:

\begin{align*}3 \sin^2 \theta + 8 \sin \theta - 3 = 0\end{align*}

Can you solve it? You might think it looks familiar.. almost like a quadratic equation, except the "x" has been replaced with a trig function? As it turns out, you're right on track. Read this Concept, and by the end, you'll be able to use the quadratic equation to solve for values of theta that satisfy the equation shown above.

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Solving Trigonometric Equations Using the Quadratic Formula

Guidance

When solving quadratic equations that do not factor, the quadratic formula is often used.

Remember that the quadratic equation is:

\begin{align*}ax^2 + bx + c = 0\end{align*} (where a, b, and c are constants)

In this situation, you can use the quadratic formula to find out what values of "x" satisfy the equation.

The same method can be applied when solving trigonometric equations that do not factor. The values for \begin{align*}a\end{align*} is the numerical coefficient of the function's squared term, \begin{align*}b\end{align*} is the numerical coefficient of the function term that is to the first power and \begin{align*}c\end{align*} is a constant. The formula will result in two answers and both will have to be evaluated within the designated interval.

Example A

Solve \begin{align*}3\ \cot^2x - 3\ \cot x = 1\end{align*} for exact values of \begin{align*}x\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solution:

\begin{align*}3 \cot^2x - 3 \cot x&= 1 \\ 3 \cot^2x - 3 \cot x - 1 & = 0\end{align*}

The equation will not factor. Use the quadratic formula for \begin{align*}\cot x\end{align*}, \begin{align*}a =3, b = -3, c = -1\end{align*}.

\begin{align*}\cot x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \cot x & = \frac{- (-3) \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)}\\ \cot x & = \frac{3 \pm \sqrt{9 + 12}}{6} \\ \cot x & = \frac{3+ \sqrt{21}}{6} \qquad \qquad \qquad \quad \text{or} \quad && \cot x = \frac{3- \sqrt{21}}{6} \\ \cot x & = \frac{3+4.5826}{6} && \cot x = \frac{3-4.5826}{6} \\ \cot x & = 1.2638 && \cot x =-0.2638 \\ \tan x & = \frac{1}{1.2638} && \tan x = \frac{1}{-0.2638} \\ x & = 0.6694, 3.81099 && x = 1.8287, 4.9703\end{align*}

Example B

Solve \begin{align*}-5 \cos^2x + 9 \sin x + 3 = 0\end{align*} for values of \begin{align*}x\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solution: Change \begin{align*}\cos^2 x\end{align*} to \begin{align*}1 - \sin^2 x\end{align*} from the Pythagorean Identity.

\begin{align*}-5 \cos^2x + 9 \sin x + 3 & = 0 \\ -5 (1 - \sin^2x) + 9 \sin x + 3 & = 0 \\ -5 + 5 \sin^2x + 9 \sin x + 3 & = 0 \\ 5 \sin^2x + 9 \sin x - 2 & = 0\end{align*}

\begin{align*}& \sin x = \frac{- 9 \pm \sqrt{9^2 - 4(5)(-2)}}{2(5)} \\ & \sin x = \frac{-9 \pm \sqrt{81+40}}{10} \\ & \sin x = \frac{-9 \pm \sqrt{121}}{10} \\ & \sin x = \frac{-9 + 11}{10} \ \text{and}\ \sin x = \frac{-9-11}{10} \\ & \sin x = \frac{1}{5}\ \text{and}\ -2 \\ &\sin^{-1} (0.2)\ \text{and}\ \sin^{-1}(-2) \end{align*}

\begin{align*}x \approx .201\ rad\end{align*} and \begin{align*}\pi -.201 \approx 2.941\end{align*}

This is the only solutions for \begin{align*}x\end{align*} since \begin{align*}-2\end{align*} is not in the range of values.

Example C

Solve \begin{align*}3 \sin^2x - 6 \sin x - 2 = 0\end{align*} for values of \begin{align*}x\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solution: \begin{align*} 3 \sin^2x - 6 \sin x - 2 & = 0\end{align*}

\begin{align*}& \sin x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(-2)}}{2(3)} \\ & \sin x = \frac{6 \pm \sqrt{36-24}}{6} \\ & \sin x = \frac{6 \pm \sqrt{12}}{6} \\ & \sin x = \frac{6 + 3.46}{10} \ \text{and}\ \sin x = \frac{6-3.46}{10} \\ & \sin x = .946\ \text{and}\ .254 \\ &\sin^{-1} (0.946)\ \text{and}\ \sin^{-1}(0.254) \end{align*}

\begin{align*}x \approx 71.08\ deg\end{align*} and \begin{align*}\approx 14.71\ deg\end{align*}

Vocabulary

Quadratic Equation: A quadratic equation is an equation of the form \begin{align*}ax^2 + bx + c = 0\end{align*}, where \begin{align*}a\end{align*}, \begin{align*}b\end{align*}, and \begin{align*}c\end{align*} are real constants.

Guided Practice

1. Solve \begin{align*}\sin^2 x - 2 \sin x - 3 = 0\end{align*} for \begin{align*}x\end{align*} over \begin{align*}[0, \pi]\end{align*}.

2. Solve \begin{align*}\tan^2x + \tan x - 2 = 0\end{align*} for values of \begin{align*}x\end{align*} over the interval \begin{align*}\left [- \frac{\pi}{2},\frac{\pi}{2} \right ]\end{align*}.

3. Solve the trigonometric equation such that \begin{align*}5 \cos^2 \theta - 6 \sin \theta = 0\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solutions:

1. You can factor this one like a quadratic.

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

For this problem the only solution is \begin{align*}\frac{3\pi}{2}\end{align*} because sine cannot be \begin{align*}3\end{align*} (it is not in the range).

2. \begin{align*}\tan^2 x + \tan x - 2 = 0\end{align*}

\begin{align*}\frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2} & = \tan x \\ \frac{-1 \pm \sqrt{1 + 8}}{2} & = \tan x \\ \frac{-1 \pm 3}{2} & = \tan x \\ \tan x & = -2 \quad \text{or}\quad 1\end{align*}

\begin{align*}\tan x = 1\end{align*} when \begin{align*}x = \frac{\pi}{4}\end{align*}, in the interval \begin{align*}\left [-\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}

\begin{align*}\tan x = -2\end{align*} when \begin{align*}x = -1.107 \ rad\end{align*}

3. \begin{align*}5 \cos^2 \theta - 6 \sin \theta = 0\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

\begin{align*}5 \left(1 - \sin^2 x \right ) - 6 \sin x & =0 \\ -5 \sin^2 x - 6 \sin x + 5 & = 0 \\ 5 \sin^2 x + 6 \sin x - 5 & = 0 \\ \frac{-6 \pm \sqrt{6^2 - 4(5)(-5)}}{2(5)} & = \sin x \\ \frac{-6 \pm \sqrt{36 + 100}}{10} & = \sin x \\ \frac{-6 \pm \sqrt{136}}{10} & = \sin x \\ \frac{-6 \pm 2 \sqrt{34}}{10} & = \sin x \\ \frac{-3 \pm \sqrt{34}}{5} & = \sin x\end{align*}

\begin{align*}x = \sin^{-1} \left (\frac{-3 + \sqrt{34}}{5} \right )\end{align*} or \begin{align*}\sin^{-1}\left (\frac{-3 - \sqrt{34}}{5} \right )\end{align*} \begin{align*}x = 0.6018\ rad\end{align*} or \begin{align*}2.5398\ rad\end{align*} from the first expression, the second expression will not yield any answers because it is out the the range of sine.

Concept Problem Solution

The original equation to solve was:

\begin{align*}3 \sin^2 \theta + 8 \sin \theta - 3 = 0\end{align*}

Using the quadratic formula, with \begin{align*}a=3, b=8, c=-3\end{align*}, we get:

\begin{align*}\sin \theta = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}= \frac{-8 \pm \sqrt{64 - (4)(3)(-3)}}{6}= \frac{-8 \pm \sqrt{100}}{6} = \frac{-8 \pm 10}{6} = \frac{1}{3} or -3 \end{align*}

The solution of -3 is ignored because sine can't take that value, however:

\begin{align*}\sin^{-1} \frac{1}{3}= 19.471^\circ\end{align*}

Practice

Solve each equation using the quadratic formula.

  1. \begin{align*}3x^2+10x+2=0\end{align*}
  2. \begin{align*}5x^2+10x+2=0\end{align*}
  3. \begin{align*}2x^2+6x-5=0\end{align*}

Use the quadratic formula to solve each quadratic equation over the interval \begin{align*}[0,2\pi)\end{align*}.

  1. \begin{align*}3\cos^2(x)+10\cos(x)+2=0\end{align*}
  2. \begin{align*}5\sin^2(x)+10\sin(x)+2=0\end{align*}
  3. \begin{align*}2\sin^2(x)+6\sin(x)-5=0\end{align*}
  4. \begin{align*}6\cos^2(x)-5\cos(x)-21=0\end{align*}
  5. \begin{align*}9\tan^2(x)-42\tan(x)+49=0\end{align*}
  6. \begin{align*}\sin^2(x)+3\sin(x)=5\end{align*}
  7. \begin{align*}3\cos^2(x)-4\sin(x)=0\end{align*}
  8. \begin{align*}-2\cos^2(x)+4\sin(x)=0\end{align*}
  9. \begin{align*}\tan^2(x)+\tan(x)=3\end{align*}
  10. \begin{align*}\cot^2(x)+5\tan(x)+14=0\end{align*}
  11. \begin{align*}\sin^2(x)+\sin(x)=1\end{align*}
  12. What type of sine or cosine equations have no solution?

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Vocabulary

Quadratic Equation

A quadratic equation is an equation that can be written in the form =ax^2 + bx + c = 0, where a, b, and c are real constants and a\ne 0.

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Sep 26, 2012
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