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) 4cos^{2} x - 1 = 0

Isolate for the
square term

4cos^{2} x = 1

divide by 4

cos^{2} 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) 2sin^{2} 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) 3cos^{2} 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) 2cos^{2} 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 - sin^{2} x) - sin x - 1 = 0

distribute

2 - 2sin^{2} x - sin x - 1 = 0

Move to the other
side to make the quadratic term positive

0 = 2sin^{2} 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 0^{o}< q<
360^{o}

Make sure your
calculator is set to degree mode!!

a) tan^{2}q -
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=
71.6^{o} or 63.6^{o}

These are the
reference angles. We also get answers in quad III where the
tangent is also positive.

q=
180 + 71.6 = 251.6^{o}

q=
180 + 63.6 = 243.6^{o}

Thus, the answers are:^{ }71.6^{o},
63.6^{o}, 251.6^{o}, 243.6^{o}

b) tan^{2} q
- 2sec^{2} q +
4 = 0

Use pythagorean
relationship to get equation in the same trig function.

(sec^{2} q
- 1)
- 2sec^{2} q + 4 = 0

-sec^{2} q
+ 3
= 0

-sec^{2} q =
-3

sec^{2} q =
3
__

sec q = + \/ 3

q = 54.7^{o}

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 + cot^{2}x)/cot x

a) sin^{2}x

b) cos^{2}x

c) sec^{2}x

d) csc^{2}x

e) tan^{2}x

5) Solve for 0 < x < 2p.
(sin x)(tan x) = sin x Give answers to the nearest
hundredth
of a radian.