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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: $2\cos^2 x-3 \cos x + 1=0$ . If $x$ falls in the interval $0 \le x < 2\pi$ , 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 $\sin^2x-3 \sin x+2=0$ .

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

#### Example B

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

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

$1- \sin x&=\sqrt{3} \cdot \sqrt{1- \sin^2x} \\(1- \sin x)^2&=\sqrt{3-3 \sin^2x}^2 \\1-2 \sin x+\sin^2x&=3-3 \sin^2x \\4 \sin^2x-2 \sin x-2&=0 \\2 \sin^2x-\sin x-1&=0 \\(2 \sin x+1)(\sin x-1)&=0$

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

#### Example C

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

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

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

$x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}$ and $x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}$

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

Concept Problem Revisit We can solve this problem by factoring.

$2\cos^2 x -3 \cos x + 1=0\\(2\cos x - 1)(\cos x - 1) = 0\\\cos x = \frac{1}{2} \text OR x = 1$ .

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

### Guided Practice

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

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

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

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

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

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

$\cos x&=0 \qquad \qquad \ \ \sin x=1 \qquad \ \ \sin x=-1 \\x&=\frac{\pi}{2} \ and \ \frac{3 \pi}{2} \qquad \ x=\frac{\pi}{2} \qquad \qquad x=\frac{3 \pi}{2}$

2. Recall that $\sin(-x)=- \sin x$ 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 $x \approx 2 \pi - 0.6283 \approx 5.6549$ and $x \approx \pi + 1.2570 \approx 4.3982$ .

### Practice

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

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

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

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