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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.

sin2x3sinx+2(sinx2)(sinx1)sinx=2 and sinx=0=0=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

1sinx(1sinx)212sinx+sin2x4sin2x2sinx22sin2xsinx1(2sinx+1)(sinx1)=31sin2x=33sin2x2=33sin2x=0=0=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.

tanx=5±(5)24(1)(9)2=5±6126.41 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.

sin2xcosxcosxcosx(sin2x1)cosx(sinx1)(sinx+1)=0=0=0

cosxx=0  sinx=1  sinx=1=π2 and 3π2 x=π2x=3π2

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

sin2xsin2xsin2x+2sinx+1(sinx+1)2sinxx=2sin(x)+1=2sinx+1=0=0=1=3π2

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

cosx=2±224(4)(1)2(4)=2±208=1±254

xcos1(1+54)cos10.8090and0.6283xcos1(154)  cos10.3090  1.8850 (reference angle is π1.88501.2570)

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. yy=sin2x=3sinx1
  1. yy=4cosx3=2tanx

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