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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 ]

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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