Transcript 10-4

Solving Trigonometric Equations
Section 10-4
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• Acceleration due to gravity: g = 9.8 m/s2. In physics, you learn
that this is a constant, but in actuality, this value changes slightly
based on your latitude.
• You can approximate g using the following formula, where 
represents the degrees of latitude:
g  9.78049(1 + 0.005288 sin2  - 0.000006 sin2 2)
• What is g in Chicago, which has a latitude of 42N?  9.81504
• Does g get larger or smaller as you increase your latitude? Why?
Methods for Solving the
Trigonometric Equation f(x) = g(x)
Graphing
a) Use a graphing calculator and graph y1 = f(x) and y2 = g(x)
b) Use 2nd Calc Intersect to find the points of intersection
Algebraically (usually the method of choice)
a) Draw a quick sketch of the graphs y = f(x) and y = g(x)
so you know how many solutions to look for.
b) If the equation involves functions of 2x AND x, transform
the functions of 2x into functions of x by using identities.
c) If the equation involves functions of 2x ONLY, then it’s
usually better to solve for 2x directly and then find x.
d) Be careful not to lose (or gain) any roots!
Let’s review a little bit of what we used to do:
0  x  2
1) Solve 2cos2x - sinx = 1 for
x
 5 3
6
,
6
,
2
0  x  2
2) Solve 2cos2x = cosx for
x
 3  5
2
,
2
,
3
,
3
3) Solve √3cos2x = 3sin2x for
x  15, 105, 195, 285
0  x  360
Solve
cos2x = 1 - sinx
for 0 ≤ x < 2π
2
1
y = 1 - sinx
Let’s graph the functions to
see what we’re working with.

2
-1
š
2
3
2
y = cos2x
There are four points of intersection in the range we are given.
Solve
cos2x = 1 - sinx
y = 1 - sinx
2
1

2
-1
š
2
3
2
y = cos2x
for 0 ≤ x < 2π
Now that we know how many
points to look for, let’s solve
this problem algebraically.
The equation contains a
function of 2x and a
function of just x, so we
should convert the 2x
function into something
with functions of just x.
So obviously we have to change cos2x into something else…
but which version of cos2x should we use?
cos2x = 1 - 2sin2x
Solve
cos2x < 1 - sinx
y = 1 - sinx
2
1

2
2
3
2
š
for 0 ≤ x < 2π
Now that we know where the
two graphs intersect, we can
study the positions of the
graphs to determine when
cos2x is less than 1 - sinx.
y = cos2x
-1
What does that physically
look like on a graph, for
one function to be “less”
than another?
So we’re looking for when cos2x is below the graph of 1 - sinx.
That happens when:

6
x
5
6
and   x  2 .
Solve
3cos2x + cosx = 2
for 0 ≤ x < 2π
What method would we have to use here to solve this equation
algebraically?
Turn cos2x into 2cos2x - 1
So now we have 3(2cos2x - 1) + cosx = 2, and you can solve that!
4
3
y = 2 - cosx
2
1

-1
y = 3cos2x
-2
-3
2π
Solve
2sin2x = 1
for 0 ≤ x < 360º
What method would we have to use here to solve this equation
algebraically?
Since we only have functions of 2x, and no
functions of x, we can solve directly for 2x and
then find x using what we know about the graph
and the period of the function y = sin2x.
This isn’t anything new. We solved
equations like this in Chapter 8.
In fact, #3 on your Chapter 8 test
was to solve 2sin2x - 1 = 0.
2
1

So… what’s x?
-1
-2
2π
Find the smallest positive root of sin2.8x = cosx
Can you think of a way to solve this problem algebraically?
We’re actually MUCH better off solving this problem on a
graphing calculator.
y = cosx
1

-1
y = sin2.8x
2π
Try these problems now:
1) cos2x - sinx > 0
2) cos2x = cosx
3) cos2x < 6cosx - 9