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Trigonometric Equations Using Factoring

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Solving Trigonometric Equations Using Quadratic Techniques
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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

Answers

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}

3. This quadratic is not factorable, so use the quadratic formula.

\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

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