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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: 2cos2x3cosx+1=0$2\cos^2 x-3 \cos x + 1=0$ . If x$x$ falls in the interval 0x<2π$0 \le x < 2\pi$ , what is/are its possible value(s)?

### Guidance

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 sin2x3sinx+2=0$\sin^2x-3 \sin x+2=0$ .

Solution: This sine equation looks a lot like the quadratic x23x+2=0$x^2-3x+2=0$ which factors to be (x2)(x1)=0$(x-2)(x-1)=0$ and the solutions are x=2$x = 2$ and 1. We can factor the trig equation in the exact same manner. Instead of just x$x$ , we will have sinx$\sin x$ in the factors.

\sin^2x-3 \sin x+2&=0 \\
(\sin x-2)(\sin x-1)&=0 \\
\sin x=2 \ and \ \sin x&=1

There is no solution for sinx=2$\sin x=2$ and sinx=1$\sin x=1$ when x=π2±2πn$x= \frac{\pi}{2} \pm 2 \pi n$ .

#### Example B

Solve 1sinx=3cosx$1- \sin x=\sqrt{3} \cos x$ in the interval 0x<2π$0 \le x < 2 \pi$ .

Solution: To solve this equation, use the Pythagorean Identity sin2x+cos2x=1$\sin^2x+\cos^2x=1$ . Solve for either cosine and substitute into the equation. cosx=1sin2x$\cos x=\sqrt{1- \sin^2 x}$

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

Solving each factor for x$x$ , we get sinx=12x=7π6$\sin x=- \frac{1}{2} \rightarrow x=\frac{7 \pi}{6}$ and 11π6$\frac{11 \pi}{6}$ and sinx=1x=π2$\sin x=1 \rightarrow x=\frac{\pi}{2}$ .

#### Example C

Solve tan2x5tanx9=0$\tan^2x-5 \tan x-9=0$ in the interval 0x<π$0 \le x < \pi$ .

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

\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

xtan16.411.416 rad$x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}$ and xtan11.410.954 rad$x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}$

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, π0.954=2.186 rad$\pi - 0.954=2.186 \ \text{rad}$ .

Concept Problem Revisit We can solve this problem by factoring.

2cos2x3cosx+1=0(2cosx1)(cosx1)=0cosx=12ORx=1
.

Over the interval 0x<2π$0 \le x < 2\pi$ , cosx=12$\cos x = \frac{1}{2}$ when x=π3$x = \frac{\pi}{3}$ and cosx=1$\cos x = 1$ when x=0$x = 0$ .

### Guided Practice

Solve the following trig equations using any method in the interval 0x<2π$0 \le x < 2 \pi$ .

1. sin2xcosx=cosx$\sin^2x \cos x=\cos x$

2. sin2x=2sin(x)+1$\sin^2x=2 \sin(-x)+1$

3. 4cos2x2cosx1=0$4 \cos^2x-2 \cos x-1=0$

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

\sin^2x \cos x- \cos x&=0 \\
\cos x(\sin^2x-1)&=0 \\
\cos x(\sin x -1)(\sin x+1)&=0

2. Recall that sin(x)=sinx$\sin(-x)=- \sin x$ from the Negative Angle Identities.

\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}

\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}

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)

The other solutions in the range are x2π0.62835.6549$x \approx 2 \pi - 0.6283 \approx 5.6549$ and xπ+1.25704.3982$x \approx \pi + 1.2570 \approx 4.3982$ .

### Explore More

Solve the following trig equations using any method. Find all solutions in the interval 0x<2π$0 \le x < 2 \pi$ . Round any decimal answers to 4 decimal places.

1. 2cos2xsinx1=0$2 \cos^2x-\sin x -1=0$
2. 4sin2x+5sinx+1=0$4 \sin^2x+5 \sin x+1=0$
3. 3tan2xtanx=0$3 \tan^2x- \tan x=0$
4. 2cos2x+cos(x)1=0$2 \cos^2x+\cos(-x)-1=0$
5. 1sinx=2cosx$1- \sin x=\sqrt{2} \cos x$
6. sinx=2sinx1$\sqrt{\sin x}=2 \sin x-1$
7. sin3xsinx=0$\sin^3x-\sin x=0$
8. tan2x8tanx+7=0$\tan^2x-8 \tan x+7=0$
9. 5cos2x+3cosx2=0$5 \cos^2x+3 \cos x-2=0$
10. sinxsinxcos2x=1$\sin x- \sin x \cos^2x=1$
11. cos2x3cosx+2=0$\cos^2x-3 \cos x+2=0$
12. sin2xcosx=4cosx$\sin^2x \cos x=4 \cos x$
13. cosxcsc2x+2cosx=6cosx$\cos x \csc^2x+2 \cos x=6 \cos x$

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval 0x<2π$0 \le x < 2 \pi$ .

1. y&=\sin^2x \\
y&=3 \sin x-1
1. y&=4 \cos x-3 \\
y&=-2 \tan x