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Trigonometric Equations Using Factoring

Factoring and the Quadratic Formula.

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Solving Trigonometric Equations Using Quadratic Techniques

As Agent Trigonometry, you are given this clue: 2cos2x3cosx+1=0 . If x falls in the interval 0x<2π , 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 .

Solution: This sine equation looks a lot like the quadratic x23x+2=0 which factors to be (x2)(x1)=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 sinx 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 and sinx=1 when x=π2±2πn .

Example B

Solve 1sinx=3cosx in the interval 0x<2π .

Solution: To solve this equation, use the Pythagorean Identity sin2x+cos2x=1 . Solve for either cosine and substitute into the equation. cosx=1sin2x

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 sinx=12x=7π6 and 11π6 and sinx=1x=π2 .

Example C

Solve tan2x5tanx9=0 in the interval 0x<π .

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 and xtan11.410.954 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 .

Concept Problem Revisit We can solve this problem by factoring.

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

Over the interval 0x<2π , cosx=12 when x=π3 and cosx=1 when x=0 .

Guided Practice

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

1. sin2xcosx=cosx

2. sin2x=2sin(x)+1

3. 4cos2x2cosx1=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)=sinx 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 x2π0.62835.6549 and xπ+1.25704.3982 .

Explore More

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

  1. 2cos2xsinx1=0
  2. 4sin2x+5sinx+1=0
  3. 3tan2xtanx=0
  4. 2cos2x+cos(x)1=0
  5. 1sinx=2cosx
  6. sinx=2sinx1
  7. sin3xsinx=0
  8. tan2x8tanx+7=0
  9. 5cos2x+3cosx2=0
  10. sinxsinxcos2x=1
  11. cos2x3cosx+2=0
  12. sin2xcosx=4cosx
  13. cosxcsc2x+2cosx=6cosx

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

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

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