**8-1
Easier Trigonometric Equations**
** **
*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 2***pfor 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 0*^{o} __<__ q
< 360^{o}

*
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=
48.6*^{o}
*This is rounded to
tenths
place. Now find the answers in the correct quadrants.*
*In three, it's 180
+
48.6 = 228.6*^{o}
*In four, it's 360 -
48.6
= 311.4*^{o}
*Therefore, the answers
are
q= 228.6*^{o} and 311.4^{o}

* *

*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 + 2n***p and x = 3.47 + 2np*

__Inclination
and slope__
*The inclination
of a line is the angle a, where 0*^{o}
__<__ a< 180^{o}, that is
measured
from the positive x-axis to the line. Lines going uphill will
have
a slope smaller than 90^{o} and lines going downhill will have
slopes greater than 90^{o}. Study the two graphs below.
*For any line with
slope
m and inclination a,*
*m = tan a
if adoesn't = 0*
*If a=90*^{o},
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 90*^{o}.
*a=
Tan*^{-1}
(3/4)
(reference angle)
*a=
37*
*So our inclination
is
a= 180 - 37 = 143*^{o}
* *
*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 90*^{o}.
*a=
Tan*^{-1} (2/5)
*a=
22*^{o}
*So our inclination
is
22*^{o}

**That's
it for this section. On to the next section!!**