<< Chapter < Page
  Math 1508 (lecture) readings in     Page 10 / 12
Chapter >> Page >
sin 2 x = 0 , 5 sin x = ± 0 , 5

This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic.

The next level of complexity comes when we need to solve a trinomial which contains trigonometric functions. It is much easier in this case to use temporary variables . Consider solving

tan 2 ( 2 x + 1 ) + 3 tan ( 2 x + 1 ) + 2 = 0

Here we notice that tan ( 2 x + 1 ) occurs twice in the equation, hence we let y = tan ( 2 x + 1 ) and rewrite:

y 2 + 3 y + 2 = 0

That should look rather more familiar. We can immediately write down the factorised form and the solutions:

( y + 1 ) ( y + 2 ) = 0 y = - 1 OR y = - 2

Next we just substitute back for the temporary variable: tan ( 2 x + 1 ) = - 1 or tan ( 2 x + 1 ) = - 2 And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the range of the trigonometric function. In that case you need to discard that solution. For example consider the same equation with cosines instead of tangents cos 2 ( 2 x + 1 ) + 3 cos ( 2 x + 1 ) + 2 = 0 Using the same method we find that cos ( 2 x + 1 ) = - 1 or cos ( 2 x + 1 ) = - 2 The second solution cannot be valid as cosine must lie between - 1 and 1. We must, therefore, reject the second equation. Only solutions to the first equation will be valid.

More complex trigonometric equations

Here are two examples on the level of the hardest trigonometric equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything!

Solve 2 cos 2 x - cos x - 1 = 0 for x [ - 180 ; 360 ]

  1. We note that cos x occurs twice in the equation. So, let y = cos x . Then we have 2 y 2 - y - 1 = 0 Note that with practice you may be able to leave out this step.

  2. Factorising yields

    ( 2 y + 1 ) ( y - 1 ) = 0
    y = - 0 , 5 or y = 1
  3. We thus get

    cos x = - 0 , 5 or cos x = 1

    Both equations are valid ( i.e.  lie in the range of cosine). General solution:

    cos x = - 0 , 5 [ 60 ] II : x = 180 - 60 + 360 · n , n Z = 120 + 360 · n , n Z III : x = 180 + 60 + 360 · n , n Z = 240 + 360 · n , n Z
    cos x = 1 [ 90 ] I ; IV : x = 0 + 360 · n , n Z = 360 · n , n Z

    Now we find the specific solutions in the interval [ - 180 ; 360 ] . Appropriate values of n yield

    x = - 120 ; 0 ; 120 ; 240 ; 360

Solve for x in the interval [ - 360 ; 360 ] :

2 sin 2 x - sin x cos x = 0
  1. Factorising yields

    sin x ( 2 sin x - cos x ) = 0

    which gives two equations

    sin x = 0
    2 sin x = cos x 2 sin x cos x = cos x cos x 2 tan x = 1 tan x = 1 2
  2. General solution:

    sin x = 0 [ 0 ] x = 180 · n , n Z
    tan x = 1 2 [ 26 , 57 ] I ; III : x = 26 , 57 + 180 · n , n Z

    Specific solution in the interval [ - 360 ; 360 ] : x = - 360 ; - 206 , 57 ; - 180 ; - 26 , 57 ; 0 ; 26 , 57 ; 180 ; 206 , 25 ; 360

Solving trigonometric equations

    1. Find the general solution of each of the following equations. Give answers to one decimal place.
    2. Find all solutions in the interval θ [ - 180 ; 360 ] .
      1. sin θ = - 0 , 327
      2. cos θ = 0 , 231
      3. tan θ = - 1 , 375
      4. sin θ = 2 , 439
    1. Find the general solution of each of the following equations. Give answers to one decimal place.
    2. Find all solutions in the interval θ [ 0 ; 360 ] .
      1. cos θ = 0
      2. sin θ = 3 2
      3. 2 cos θ - 3 = 0
      4. tan θ = - 1
      5. 5 cos θ = - 2
      6. 3 sin θ = - 1 , 5
      7. 2 cos θ + 1 , 3 = 0
      8. 0 , 5 tan θ + 2 , 5 = 1 , 7
    1. Write down the general solution for x if tan x = - 1 , 12 .
    2. Hence determine values of x [ - 180 ; 180 ] .
    1. Write down the general solution for θ if sin θ = - 0 , 61 .
    2. Hence determine values of θ [ 0 ; 720 ] .
    1. Solve for A if sin ( A + 20 ) = 0 , 53
    2. Write down the values of A [ 0 ; 360 ]
    1. Solve for x if cos ( x + 30 ) = 0 , 32
    2. Write down the values of x [ - 180 ; 360 ]
    1. Solve for θ if sin 2 ( θ ) + 0 , 5 sin θ = 0
    2. Write down the values of θ [ 0 ; 360 ]

Questions & Answers

discuss how the following factors such as predation risk, competition and habitat structure influence animal's foraging behavior in essay form
Burnet Reply
location of cervical vertebra
KENNEDY Reply
What are acid
Sheriff Reply
define biology infour way
Happiness Reply
What are types of cell
Nansoh Reply
how can I get this book
Gatyin Reply
what is lump
Chineye Reply
what is cell
Maluak Reply
what is biology
Maluak
what's cornea?
Majak Reply
what are cell
Achol
Explain the following terms . (1) Abiotic factors in an ecosystem
Nomai Reply
Abiotic factors are non living components of ecosystem.These include physical and chemical elements like temperature,light,water,soil,air quality and oxygen etc
Qasim
Define the term Abiotic
Marial
what is biology
daniel Reply
what is diffusion
Emmanuel Reply
passive process of transport of low-molecular weight material according to its concentration gradient
AI-Robot
what is production?
Catherine
hello
Marial
Pathogens and diseases
how did the oxygen help a human being
Achol Reply
how did the nutrition help the plants
Achol Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Math 1508 (lecture) readings in precalculus' conversation and receive update notifications?

Ask