Transcript Solving Trigonometric Equations

Lesson 5.3
Solving
Trigonometric Equations
Solving
Trigonometric Equations
To solve trigonometric equations:
Use standard algebraic techniques learned in Algebra II.
Look for factoring and collecting like terms.
Isolate the trig function in the equation.
Use the inverse trig functions to assist in determining
solutions.
Solving
Trigonometric Equations
For all problems,
The solution interval
Will be
[0, 2)
You are responsible for checking your solutions back into the original problem!
Solving
Trigonometric Equations
Solve:
2cos x 1  0
Step 1: Isosolate cos x using algebraic skills.
2cos 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  2 cot x
Step 1:
cot x cos 2 x  2 cot x  0
cot x  cos 2 x  2   0
cot x  0 or cos 2 x  2  0
cos 2 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.
3 tan 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).
Solving
Trigonometric Equations
Solve:
2sin 2 x  3cos x  3  0
2 1  cos 2 x   3cos x  3  0
2  2 cos 2 x  3cos x  3  0
2 cos 2 x  3cos x  1  0
2 cos 2 x  3cos x  1  0
 2cos x 1 cos x 1  0
2 cos 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
cos 2 x  2 cos x  1  sin 2 x
FOIL or Double Distribute
cos 2 x  2 cos x  1  1  cos 2 x
2 cos 2 x  2 cos x  0
2cos x  cos x  1  0
2 cos x  0 or cos x  1  0
cos x  1
cos x  0
x
 3
,X
2 2
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
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  2 cos 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
Solving
Trigonometric Equations
What you should know:
1. How to use algebraic techniques to solve
trigonometric equations.
2. How to solve quadratic trigonometric equations
by factoring or the quadratic formula.
3. How to solve trigonometric equations involving
multiple angles.