Download 5. 1 SOLVING Trig Equations

Start Up Day 37
1. Simplify:
2, Verify:
1
-
1
cscq +1 cscq -1
sec 2 a
= sec a csc a
æp
ö
cot ç - a ÷
è2
ø
1
SOLVING TRIGONOMETRIC EQUATIONS-DAY 37
OBJECTIVE: SWBAT SOLVE
TRIGONOMETRIC EQUATIONS.
EQ: How can we use trigonometric identities to
help solve trig equations?
What is the difference between solving for all
solutions (General Solution) verses solving
within an interval like from 0 to 2π?
.
HOMELEARNING:
p. 411 #53, 57, 60, 63 &
73 ; p. 411 #51, 52, 55, 56, 58, 59, 64, 66, 69,
71 & 87
FOCUS on ALGEBRA 1st
All trigonometric equations are NOT identities. Some equations will only
be true at specific angles.
To solve a trigonometric equation, use standard algebraic techniques
such as collecting like terms and factoring.
Your preliminary goal in solving a trigonometric equation is to isolate
the trigonometric function in the equation. Sometimes it is helpful to
replace the trigonometric function with just an ordinary variable of your
choosing so that you can focus first on the algebra.
For example, to solve the equation 2 sin x = 1, we might let X=sin x &
re-write it as 2X=1 then we can see clearly that we should divide each
side by 2 to obtain X = ½ OR, by replacing our original function:
3
Back to TRIG: Where is sin x = ½?
To solve for x, note that the equation
has solutions x =  /6 and x = 5 /6 in the interval [0, 2).
4
Infinitely many solutions?
Moreover, because sin x has a period of 2, there are
infinitely many other solutions, which can be written as
and
where n is an integer..
“+ 2nπ” means “+ any number of full circles”
This is known as the “General solution”
5
Back to the “Circle of Life”
Another way to show that the equation
has
infinitely many solutions is to identify the angles on the unit
circle where this sine value occurs.
Any angles that are co-terminal with  /6 or 5 /6 will also be
solutions of the equation.
When solving trigonometric
equations, you should write
your answer(s) using exact
values rather than decimal
approximations.
Figure 5.7
6
Example 2 – Collecting Like Terms
Solve
Solution:
Begin by rewriting the equation so that sin x is isolated on
one side of the equation.
Define your temporary variable..
Let X = sin x (temporarily!)
X + 2 = -X
2X + 2 = 0
2X = - 2
X=
- 2
2
RE- Write original equation.
Add “x” to each side.
Subtract
from each side.
Divide each side by 2.
7
The basics of Solving Trig
Equations
http://www.voki.com/pickup.php?scid=8888
701&height=267&width=200
8
Trigonometry & the UNIT Circle!
cont’d
Re-write with the original trig function.
Now, Identify the angles where this sine value occurs.
Because sin x has a period of 2, first find all solutions in
the interval [0, 2). (Once around the UNIT Circle)
These solutions are x = 5 /4 and x = 7 /4. Finally, add
multiples of 2 to each of these solutions to get the general
form
and
where n is an integer.
General solution
9
Equations of Quadratic Type
10
Equations of Quadratic Type
Many trigonometric equations are of quadratic type
ax2 + bx + c = 0. Here are a couple of examples.
Quadratic in sin x
Quadratic in sec x
2 sin2 x – sin x – 1 = 0
sec2 x – 3 sec x – 2 = 0
2(sin x)2 – sin x – 1 = 0
(sec x)2 – 3(sec x) – 2 = 0
To solve equations of this type, factor the quadratic or, if
this is not possible, use the Quadratic Formula.
11
Example 3 – Factoring an Equation of Quadratic Type
Find all solutions of 2 sin2 x – sin x – 1 = 0 in the interval
[0, 2).
Solution:
Begin by treating the equation as a quadratic & let X= sin x
temporarily so that you can see that factoring is the key!
Re-Write original equation.
2 X2 – X– 1 = 0
Factor.
(2 X + 1)(X– 1) = 0
12
Example 3 – Solution
cont’d
Setting each factor equal to zero, you obtain the following
solutions in the interval [0, 2).
2 X+ 1 = 0
X= - ½
and
and
X– 1 = 0
X=1
Once the Algebra is done, bring back the original trigonometric function and
identify the angles where these values occur from 0 to 2π
We don’t need to add “2nπ” to our solutions because the solution
interval was limited to a specific interval. [0, 2).
13
YOUR TURN:
1. Solve
over the interval
2. Solve:
3.
Solve:
14
Using Inverse Functions and
Identities
15
Using trigonometric Identities and Inverse
Functions
In the next example, you will see how identities and inverse
trigonometric functions can be used to solve an equation.
16
Example 4 – Using Identities & Inverse
Functions
Solve sec2 x – 2 tan x = 4.
Solution:
sec2 x – 2 tan x = 4
Notice that 2 different trig
Functions are terms withing the original equ
x – 2 tan x – 4 = 0
Use an identity to rewrite one trig function in
Terms of the other– the Pythagorean identit
Is most appropriate for this one!
1+
tan2
tan2 x – 2 tan x – 3 = 0
X2 – 2 X– 3 = 0
(X– 3)(X+ 1) = 0
Combine like terms and re-write with a
temporary variable.
Factor.
17
Example 4 – Working on the
solution.
cont’d
Setting each factor equal to zero, you obtain two
solutions,
X= 3 and X = -1
OR
tan x = 3
tan x = –1
Now identify the angles where these values occur, if
possible.
Since a tangent value of 3 is does NOT occur at one of our special angles from the
UNIT Circle, the best way to represent this solution is by using the inverse or arc
function:
x = arctan 3
18
Example 4 – A Final Solution
cont’d
Finally, because tan x has a period of , you obtain the
general solution by adding multiples of 
x = arctan 3 + n
and
General solution
where n is an integer.
You can use a calculator to approximate the value of
arctan 3, if necessary. (radian mode)
19
Functions Involving Multiple Angles
20
Functions Involving Multiple Angles
The next example involves trigonometric functions of
multiple angles of the forms sin ku and cos ku.
To solve equations of these forms, first ignore the “ku”, rewrite with a variable of your choosing, solve the equation
algebraically for your variable, replace with the original trig
function and solve the the angle like you would normally do.
Lastly, consider these “angles” are really solutions for ku,
so we must divide your result by k.
21
Example 5 – Functions of Multiple Angles
Solve 2 cos 3t – 1 = 0.
Solution: Let X = cos 3t
2 X– 1 = 0
2 X= 1
X=½
Re-Write original equation.
Add 1 to each side.
Divide each side by 2.
Replace your original trig function..
22
Example 5 – Solution
cont’d
Now identify all the angles within the interval [0, 2), where
the cosine value is ½
You should know that this occurs at  /3 and 5 /3, so, in
general, you have
but this is only solved for “3t” –so we will need a little more
algebra
Dividing these results by 3, you obtain the general solution
and
where n is an integer.
General solution
23