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 andp, 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.
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 px
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.
