8-5 Solving More Difficult Trig Equations
 
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Most of these type of problems can be solved the same way you solve basic algebraic equations.  Also, we can use many of the identities from the previous sections.  Remember, we can solve these for a few answers or an infinite number of answers (in rads or degrees)!  So watch the directions closely.
 
1)  Solve each of the following for 0 < x < 2p
 
a)  4cos2 x - 1 = 0
Isolate for the square term
4cos2 x = 1
divide by 4
cos2 x = 1/4
take the square root
cos x = + 1/2
since we have both positive and negative answers, we will have an answer in each of the four quadrants.  Find the reference angle in rads and use the reference angle formulas.  Set calculator to rads.
x = cos-1 (1/2)
x = 1.05  This is the reference angle.
In quad II, 3.14 - 1.05 = 2.09
In quad III, 3.14 + 1.05 = 4.19
In quad IV, 6.28 - 1.05 = 5.23
Thus, x = 1.05, 2.09, 4.19, 5.23
 
 
b)  2sin2 x - sin x = 0
factor
sin x(2sin x - 1) = 0
write the two solutions
sin x = 0 or 2sin x- 1 = 0
sin x = 0 or 2sin x = 1
sin x = 0 or sin x =1/2
sin x = 0 at 0 and p, x = sin-1 (1/2) --> p/6 and 5p/6
Thus, the four answers are: x = 0, p,p/6, 5p/6
 
c)  3cos2 x + 2cos x - 1 = 0
factor!!
(3cos x -1)(cos x + 1) = 0
cos x = 1/3 or cos x = -1
x = cos-1 (1/3) or x = cos-1 (-1)
x = 1.23 or x = p
The first solution had another answer in quad IV. (Why?)
x = 6.28 - 1.23 = 5.05
Thus, x = 1.23, 5.05, p
d)  2cos2 x - sin x - 1 = 0
This is much easier to solve if the function is written in terms of the same trig function.  Using the pythagorean relationship, we can write:
2(1 - sin2 x) - sin x - 1 = 0
distribute
2 - 2sin2 x - sin x - 1 = 0
Move to the other side to make the quadratic term positive
0 = 2sin2 x + sin x -1
factor
0 = (2sin x - 1)(sin x + 1)
sin x = 1/2 or sin x = -1
x = sin-1 (1/2) or x = sin-1 (-1)
x = p/6 and 5p/6, or x = 3p/2
(These are values that should have been memorized in previous sections!)
Thus, the answers are: x = p/6 and 5p/6, or x = 3p/2
 
2)  Solve each of the following for 0o < q< 360o
 
Make sure your calculator is set to degree mode!!
a)  tan2q - 5tan q + 6 = 0
factor!
(tan q - 3)(tan q -2) = 0
tan q = 3 or tan q = 2
q = tan-1 3 or q= tan-1 2
q= 71.6o or 63.6o
These are the reference angles.  We also get answers in quad III where the tangent is also positive.
q= 180 + 71.6 = 251.6o
q= 180 + 63.6 = 243.6o
Thus, the answers are:  71.6o, 63.6o, 251.6o, 243.6o
 
b)  tan2 q - 2sec2 q + 4 = 0
Use pythagorean relationship to get equation in the same trig function.
(sec2 q - 1) - 2sec2 q + 4 = 0
-sec2 q + 3 = 0
-sec2 q = -3
                                                       sec2 q = 3                    __
sec q = + \/ 3
 q = 54.7o
This is the reference angle.  Since it is both positive and negative, we have an answer in all four quadrants.
Quad II:  180 - 54.7 = 125.3o
Quad III:  180 + 54.7 = 234.7o
Quad IV:  360 - 54.7 = 305.3o
Thus, the answers are:  54.7o , 125.3o , 234.7o , 305.3o
 
c) 
 
d)  4sin q + 3cos q = 0
This one would be difficult if we tried to use the pythagorean relationship.  Since it deals with sine and cosine rather than their squares, it would introduce radicals in the equation.  Not a good choice.  It might be smart to try another relationship.  How about a third trig function.  Sine and cosine are related to Tangent!!  If we divide the above equation by cosine, we can change the problem to tangent and have only one trig function!!
Well, that does it.  Are you ready for the sample test or do you need to review?  Pick the appropriate button!!
 
 
 
 


 
 
Current quizaroo #  8
 
1)  Solve for 0 < x < 2p.    cot x = -2.5   Give answers to the nearest hundredth of a radian. 
 
a) .38 and 3.52
b)  2.76 and 5.90
c)  1.19 and 4.33
d)  1.95 and 5.09
e)  1.19 and 1.95
 
 
 
2)   Give the amplitude and period for   y = -5cos p

          a)  amplitude = -5 and period = 2

b)  amplitude = 2 and period = 2p/5
c)  amplitude = 5 and period = 2
d)  amplitude = 5 and period = 1
e)  amplitude = -5 and period = p
 

 
 
3)  Find the maximum point in the first cycle for the equation:  y - 3 = 4sin (px/2)
 
a)  (1, 7)
b)  (0, 3)
c)  (2, 8)
d)  (3, 10)
e)  (2, 7)
 
 
 
4)  Simplify  tan x( 1 + cot2x)/cot x
 
a)  sin2x
b)  cos2x
c)  sec2x
d)  csc2x
e)  tan2x
 
 
  5)   Solve for 0 < x < 2p.    (sin x)(tan x) = sin x   Give answers to the nearest hundredth of a radian.
 
a)   p/4, 5p/4
b)   0, p
c)   0, p, 2p
d)   0, 2p
e)   0, p/4, p, 5p/4
 
 

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