Download Using the Real Number Line

Bellwork
1.
2.
3.
4.
Verify
tan   cot  csc sec
What is the complement of
. Find  : tan   2
Solve ABC


3
8?
A
7
C
5.
11
B
What are the Quotient Identities?
Precalculus
5.3 Solving Trigonometric
Equations
1
Precalculus
Chapter 5.3: Solving
Trigonometric Equations
Objectives:
• Use standard algebraic techniques to solve
trig equations
•Use inverse trig functions to solve trig
equations
Example 1
• Find all solutions of
cotx + 1 = 0
• cotx = -1
• Think about inverse trig. Cotangent, of what
angle (x) gives you a ratio of -1?
• x = 3
4
Precalculus
 n
5.3 Solving Trigonometric
Equations
3
Example 2
• Solve the equation for
0≤ x <2π
4cos2x – 3 = 0
• 4cos2x = 3
• cos2x = ¾
x
3
cos x 
2
• Cosine of what angle is
Precalculus
3
2
5.3 Solving Trigonometric
Equations

6
 n
11
x
 n
6
4
Example 3
• Find all solutions in
the interval [0,2π)
sin x  2   sin x
Precalculus
5.3 Solving Trigonometric
Equations
5
Example 4
• Find all solutions in the interval [0,2π)
cos3x = cosx
cos3x - cosx = 0
cosx (cos2-1) = 0
(cosx)(- sin2x) = 0
cosx = 0  x = π/2, 3π/2
-sin2x = 0  x = 0, π
general solution:
x = 2nπ, π/2+2nπ, π + 2nπ, 3π/2+ 2nπ
Precalculus
5.3 Solving Trigonometric
Equations
6
Example 4 (cont.)
• Find all solutions in the interval [0,2π)
cos3x = cosx
what if we divided both sides by cosx?
cos2x = 1
cos2x-1 = 0
-sin2x = 0
sinx = 0  x = 0, π
we lost two solutions! x = π/2 & 3π/2
Precalculus
5.3 Solving Trigonometric
Equations
7
Example 5
• Find all solutions in the interval [0,2π)
tan3x = 1
solution:
x = π/12, 5π/12, 13 π/12, 17 π/12, 7 π/4
general solution:
x = π/12+2nπ/3 and 5π/12 + 2n π/3
Precalculus
5.3 Solving Trigonometric
Equations
8
Example 6
• Find all solutions in the interval [0,2π)
secx cscx = 2cscx
solution: x = π/3, 5π/3
general solution:
x = π/3+2nπ and x = 5π/3 + 2nπ
Precalculus
5.3 Solving Trigonometric
Equations
9
Example 7
• Find all solutions in the interval [0,2π)
2sin2x + 3sinx + 1= 0
solution: x = 7π/6, 3π/2, 11π/6
general solution:
x = 3π/2+2nπ, x = 7π/6 + 2nπ,
and x =11π/6 + 2nπ
Precalculus
5.3 Solving Trigonometric
Equations
10
Example 8
• Find all solutions in the interval [0,2π)
cosx + 1 = sinx
solution: x = π, π/2
note: 3π/2 is extraneous
general solution:
x = π + 2nπ, x = π/2 + 2nπ
Precalculus
5.3 Solving Trigonometric
Equations
11
Example 9
• Find the x-intercepts of the graph y
= sinπx + cosπx
0 = sinπx + cosπx
-sinπx = cosπx
x = -1/4, 3/4, 7/4, 11/4
General solution:
x=-1/4 +k
Precalculus
5.3 Solving Trigonometric
Equations
12
Closure
• What is an extraneous solution? Where
does it come from?
– An extraneous solution is not a valid
solution of the original equation. It is
usually the result of raising both sides of an
equation to the same power.
Precalculus
5.3 Solving Trigonometric
Equations
13
Assignment
• Assignment CW 5.3
• Pg.400: #3-71 odd
Precalculus
5.3 Solving Trigonometric
Equations
14
Bellwork
1. Solve 2x2+5=41
2. Use the quadratic formula to solve
x2-2x-4=0
3. Solve by factoring
x2-3x-4=0
Solutions
1. x=-3√2, 3√2
2. x=1+√5, 1-√5
3. x=-1, 4
Precalculus
5.3 Solving Trigonometric
Equations
15
Bellwork
Find all the solutions in the interval [0,2π)
1. 2cos x  1  0
2. 2sin 2x  1
2
Precalculus
5.3 Solving Trigonometric
Equations
16