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 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!!