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# 3.5: Trigonometric Equations Using the Quadratic Formula

Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometric Equations Using the Quadratic Formula
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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:

3sin2θ+8sinθ3=0\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.

### Watch This

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:

ax2+bx+c=0\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 a\begin{align*}a\end{align*} is the numerical coefficient of the function's squared term, b\begin{align*}b\end{align*} is the numerical coefficient of the function term that is to the first power and c\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 3 cot2x3 cotx=1\begin{align*}3\ \cot^2x - 3\ \cot x = 1\end{align*} for exact values of x\begin{align*}x\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.

Solution:

3cot2x3cotx3cot2x3cotx1=1=0

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

cotxcotxcotxcotxcotxcotxtanxx=b±b24ac2a=(3)±(3)24(3)(1)2(3)=3±9+126=3+216or=3+4.58266=1.2638=11.2638=0.6694,3.81099cotx=3216cotx=34.58266cotx=0.2638tanx=10.2638x=1.8287,4.9703

#### Example B

Solve 5cos2x+9sinx+3=0\begin{align*}-5 \cos^2x + 9 \sin x + 3 = 0\end{align*} for values of x\begin{align*}x\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.

Solution: Change cos2x\begin{align*}\cos^2 x\end{align*} to 1sin2x\begin{align*}1 - \sin^2 x\end{align*} from the Pythagorean Identity.

5cos2x+9sinx+35(1sin2x)+9sinx+35+5sin2x+9sinx+35sin2x+9sinx2=0=0=0=0

sinx=9±924(5)(2)2(5)sinx=9±81+4010sinx=9±12110sinx=9+1110 and sinx=91110sinx=15 and 2sin1(0.2) and sin1(2)

x.201 rad\begin{align*}x \approx .201\ rad\end{align*} and π.2012.941\begin{align*}\pi -.201 \approx 2.941\end{align*}

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

#### Example C

Solve 3sin2x6sinx2=0\begin{align*}3 \sin^2x - 6 \sin x - 2 = 0\end{align*} for values of x\begin{align*}x\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.

Solution:

3sin2x6sinx2=0

sinx=6±(6)24(3)(2)2(3)sinx=6±36246sinx=6±126sinx=6+3.4610 and sinx=63.4610sinx=.946 and .254sin1(0.946) and sin1(0.254)

x71.08 deg\begin{align*}x \approx 71.08\ deg\end{align*} and 14.71 deg\begin{align*}\approx 14.71\ deg\end{align*}

### Guided Practice

1. Solve sin2x2sinx3=0\begin{align*}\sin^2 x - 2 \sin x - 3 = 0\end{align*} for x\begin{align*}x\end{align*} over [0,π]\begin{align*}[0, \pi]\end{align*}.

2. Solve tan2x+tanx2=0\begin{align*}\tan^2x + \tan x - 2 = 0\end{align*} for values of x\begin{align*}x\end{align*} over the interval [π2,π2]\begin{align*}\left [- \frac{\pi}{2},\frac{\pi}{2} \right ]\end{align*}.

3. Solve the trigonometric equation such that 5cos2θ6sinθ=0\begin{align*}5 \cos^2 \theta - 6 \sin \theta = 0\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.

Solutions:

1. You can factor this one like a quadratic.

sin2x2sinx3=0 (sinx3)(sinx+1)=0sinx3=0  sinx=3or x=sin1(3) sinx+1=0sinx=1 x=3π2

For this problem the only solution is 3π2\begin{align*}\frac{3\pi}{2}\end{align*} because sine cannot be 3\begin{align*}3\end{align*} (it is not in the range).

2. tan2x+tanx2=0\begin{align*}\tan^2 x + \tan x - 2 = 0\end{align*}

1±124(1)(2)21±1+821±32tanx=tanx=tanx=tanx=2or1

tanx=1\begin{align*}\tan x = 1\end{align*} when x=π4\begin{align*}x = \frac{\pi}{4}\end{align*}, in the interval [π2,π2]\begin{align*}\left [-\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}

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

3. 5cos2θ6sinθ=0\begin{align*}5 \cos^2 \theta - 6 \sin \theta = 0\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.

5(1sin2x)6sinx5sin2x6sinx+55sin2x+6sinx56±624(5)(5)2(5)6±36+100106±136106±234103±345=0=0=0=sinx=sinx=sinx=sinx=sinx

\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 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*}

### Explore More

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?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 3.5.

### Vocabulary Language: English

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 $a\ne 0$.

## Date Created:

Sep 26, 2012

Feb 26, 2015
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