Transcript Solving Trigonometric Equations

Solving
Trigonometric Equations
Solving
Trigonometric Equations
For most problems,
The solution interval
Will be
[0, 2)
You are responsible for checking your solutions back into the original problem!
First Degree Trigonometric
Equations:
• These are equations where there is one
kind of trig function in the equation and
that function is raised to the first power.
2 sin(x)  1
2 sin(x)  1
1
sin( x) 
2
Now figure out where sin = -1/2 on the unit circle.
1
7
11
sin 
at
and
2
6
6
Complete the List of
Solutions:
• If you are not restricted to a specific
interval and are asked to give the general
solutions then remember that adding on
any integer multiple of 2π represents a coterminal angle with the equivalent
trigonometric ratio.
Solutions:
 7

2
k

 6
x
11


 2k
 6
Where k is an integer and gives all
the coterminal angles of the
solution.
Practice
• Solve the equation. Find the general solutions
3 csc   2  0
3 csc   2
2
csc  
3

3
which means that sin 
2
2
   2 k ,
 2 k
3
3
Solving
Trigonometric Equations
Solve:
2 cos x  1  0
Step 1: Isosolate cos x using algebraic skills.
2 cos x  1
cos x  1
2
Step 2: Determine in which quadrants cosine is positive. Use the inverse
function to assist by finding the angle in Quad I first. Then use that angle
as the reference angle for the other quadrant(s).
QI
x
QIV
 5
3
,
3
Note: cosine is positive in
Quad I and Quad IV.
Note: The reference angle is /3.
Solving
Trigonometric Equations
Solve:
Step 1:
tan x  1  0
2
tan x  1
2
tan x   1
tan x  1
2
Step 2:
Q1
x
Note: Since there is a  , all four quadrants
hold a solution with /4 being the reference
angle.
QIV
QIII
QII
 3 5 7
4
,
4
,
4
,
4
Solving
Trigonometric Equations
2
Solve: cot x cos x  2cot x
Step 1:
cot x cos2 x  2cot x  0
cot x  cos 2 x  2   0
cot x  0 or cos2 x  2  0
cos2 x  2
cos2 x   2
cos x   2
Step 2:
x
 3
2
,
2
x
Note: There is no solution here because 2
lies outside the range for cosine.
Solving
Trigonometric Equations
Try these:
Solution
3 7
,
4 4
1.
tan x  1  0
x
2.
sec x  4  0
x
3.
3tan x  tan x
x  0,
2
3
 2 4 5
3
,
3
,
 5
6
,
6
3
,
, ,
3
7 11
,
6
6
Solving
Trigonometric Equations
Solve:
2sin 2 x  sin x  1  0
 2sin x 1sin x 1  0
Factor the quadratic equation.
2sin x  1  0 or sin x 1  0
Set each factor equal to zero.
1
sin x  
2
7 11
x
,
6
6
sin x  1
x

2
Solve for sin x
Determine the correct quadrants
for the solution(s).
Solve : 4sin 2 ( x)  1  0 over the int erval [0, 2 )
This is a difference of squares and can factor
(2sin x  1)(2sin x 1)  0
Solve each factor and you should end up with 4 solutions
1
1
sin x 
and sin x 
2
2
x
 5 7 11
6
,
6
,
6
,
6
Practice
Find the general solutions for
tan x  2 tan x  1
2
tan x  2 tan x  1  0
2
(tan x  1)(tan x  1)  0
tan x  1
3
7
x
 k ,
 k
4
4
Writing in terms of 1 trig fnc
• If there is more than one trig function
involved in the problem, then use your
identities.
• Replace one of the trig functions with an
identity so there is only one trig function
being used
Solve the following
2cos x  sin x  1  0
2
Replace cos2 with 1 - sin2
2(1  sin 2 x)  sin x  1  0
2  2sin 2 x  sin x  1  0
2sin 2 x  sin x  1  0
2sin 2 x  sin x  1  0
(2sin x  1)(sin x  1)  0
1
sin x 
and sin x  1
2
7
11

x
 2k , x 
 2k , x   2k
6
6
2
Solving
Trigonometric Equations
Solve:
2sin 2 x  3cos x  3  0
2 1  cos 2 x   3cos x  3  0
2  2cos2 x  3cos x  3  0
2cos2 x  3cos x  1  0
2cos2 x  3cos x  1  0
 2cos x 1cos x 1  0
2cos x  1  0 or cos x 1  0
cos x 
x
1
2
 5
3
,
3
Replace sin2x with 1-cos2x
Distribute
Combine like terms.
Multiply through by – 1.
Factor.
Set each factor equal to zero.
cos x  1
Solve for cos x.
x0
Determine the solution(s).
Solving
Trigonometric Equations
Solve:
cos x  1  sin x
 cos x  1
2
  sin x 
Square both sides of the equation
in order to change sine into terms
of cosine giving only one trig
function to work with.
2
cos2 x  2cos x  1  sin 2 x
FOIL or Double Distribute
cos2 x  2cos x  1  1  cos2 x
2cos2 x  2cos x  0
2cos x  cos x 1  0
2cos x  0 or cos x  1  0
cos x  1
cos x  0
x
Replace sin2x with 1 – cos2x
Set equation equal to zero since it is a
quadratic equation.
Factor
Set each factor equal to zero.
Solve for cos x
 3
2
,
X2 x  
Why is 3/2 removed as a solution?
Determine the solution(s).
It is removed because it does not
check in the original equation.
Solving
Trigonometric Equations
Solve:
Solution:
1
2
cos 3 x 
No algebraic work needs to be done because cosine is already by itself.
Remember, 3x refers to an angle and one cannot divide by 3 because it
is cos 3x which equals ½.
Since 3x refers to an angle, find the angles whose cosine value is ½.
3x 
x
 5
3
,
Now divide by 3 because it is angle equaling angle.
3
 5
9
3x 
,
Notice the solutions do not exceed 2. Therefore,
more solutions may exist.
9
 5 7 11
,
,
,
3
3 3
 5 7 11
,
,
x ,
9
9 9 9
3
Return to the step where you have 3x equaling
the two angles and find coterminal angles for
those two.
Divide those two new angles by 3.
Solving
Trigonometric Equations
 5 7 11 13 17
3x  ,
,
,
,
,
3 3 3
3
3
3
x
The solutions still do not exceed 2.
Return to 3x and find two more
coterminal angles.
 5 7 11 13 17
,
,
,
,
,
9 9
9
9
9
9
Divide those two new angles by 3.
 5 7 11 13 17 19 23 The solutions still do not exceed 2.
3x  ,
,
,
,
,
,
,
Return to 3x and find two more
3 3 3
3
3 coterminal angles.
3
3
3
x
 5 7 11 13 17 19
,
,
,
,
,
,
9
9 9
9
9
9
9
Divide those two new angles by 3.
Notice that 19/9 now exceeds 2 and
is not part of the solution.
Therefore the solution to cos 3x = ½ is
x
 5 7 11 13 17
9
,
9
,
9
,
9
,
9
,
9
Solving
Trigonometric Equations
Try these:
1.
4sin 2 x  2cos x  1
2.
csc x  cot x  1
3.
3
sin 2 x  
2
4.
x
2
cos 
2
2
Solution
x  5.4218

x
2
2 5 5 11
x
,
,
,
3 6 3 6
x

2
Find the general solutions for
sin 3x +2= 1
sin 3x  1
3
3x 
2
3
2k
2
x

3
3

2
x   k
2 3
Practice
Solve 2cos 4x  3  0
 3
cos 4 x 
2
5
7
4x 
and 4 x 
6
6
5 k
7 k
x

, x

24 2
24 2