By the nature of the trig functions, solving
trig equations will yield an infinite number of solutions. Watch
closely to the domain of x as we do each problem. Sometimes we will
ask for all possible solutions, called the general
solutions of a trig equation. These are generally found
by finding particular solutions and
adding in a factor for the period length of the particular function.
Many times we will restrict the domain of x to the first period of the
given function or one time around the unit circle, 2p.
Sample
Problems
1) Find the values of x between
0 and 2pfor which cos x = .4
Solution: To find x, we are looking
for the angle measured in rads, one time around the unit circle.
Since the cos is positive in quads 1 and 4, we should find two answers
in the given domain.
x = cos-1
.4
Use your calculator to get the answer in the
first quadrant. Round to 4 decimal places.
x = 1.1593
This is the solution in quadrant 1. The
other solution is in quad 4. Remembering our reference angles, the
other answer is found by taking 2p-1.1593 =
5.1239
Therefore, the answers are x
= 1.1593 and 5.1239 2) Solve 4 sin
q + 5 = 2 for 0o< q
< 360o
Solution:
Isolate the variable.
4 sin q
= -3
sin q=
-3/4
q=
sin-1 (-3/4)
The answers this time
are in quadrants III and IV because the sin is negative in those quadrants.
Find the reference angle first then apply the correct reference angle formula.
q=
sin-1(3/4)
q=
48.6o
This is rounded to tenths
place. Now find the answers in the correct quadrants.
In three, it's 180 +
48.6 = 228.6o
In four, it's 360 - 48.6
= 311.4o
Therefore, the answers are
q= 228.6o and 311.4o 3) Solve the equation
sec x = 4.2 for x between 0 and 2p
Solution: Solve the same way
as the first example
x = sec-1
4.2
x = 1.3304
This is the answer in the first quadrant.
Sec is also positive in quadrant IV. Using reference angles, we get
2p- 1.3304 = 4.9528
Therefore the solutions are x
= 1.3304 and 4.9528 ( Note: If we wanted
all solutions, we would simply add multiples of 2pto
each of the above answers. x = 1.3304 + 2np and
4.9528 + 2np) 4) Solve the following
equation for all solutions. Round answers to the nearest hundredth
of a radian.
Solve for x:
12 cot x = 35
cot x = 35/12
x = cot-1
(35/12)
x = .33
Since the cot is positive in Quads I and III,
we want the above answer plus this one: 3.14 + .33 = 3.47
Therefore, all answers are:
x = .33 + 2np and x = 3.47 + 2np
Inclination
and slope
The inclination
of a line is the angle a, where 0o< a< 180o, that is measured
from the positive x-axis to the line. Lines going uphill will have
a slope smaller than 90o and lines going downhill will have
slopes greater than 90o. Study the two graphs below.
For any line with slope
m and inclination a,
m = tan a
if adoesn't = 0
If a=90o,
then the line has undefined slope.
Examples
1) To the nearest degree, find the inclination
of the line 3x + 4y = 8 Solution:
Find the slope of the equation.
3x + 4y = 8
4y = -3x + 8
y = (-3/4)x + 2
m = -3/4
The line is going downhill, so the inclination
will be greater than 90o.
a=
Tan-1 (3/4)
(reference angle)
a=
37
So our inclination is
a= 180 - 37 = 143o
2) To the nearest degree,
find the inclination of the line -2x + 5y = 15
Solution:
Same as above.
-2x + 5y = 15
5y = 2x + 15
y = (2/5)x + 3
m = 2/5
The line is going uphill
because the slope is positive, so the inclination will be less than 90o.
a=
Tan-1 (2/5)
a=
22o
So our inclination is
22o
That's
it for this section. On to the next section!!
