GET OUT PAPER FOR
NOTES!!!
Warm-up (3:30 m)
1. Solve for all solutions graphically: sin3x = –cos2x
2. Molly found that the solutions to cos x = 1 are
x = 0 + 2kπ AND x = 6.283 + 2kπ, k   . Is
Molly’s solution correct? Why or why not?
sin3x = –cos2x
cos x = 1
• x = 0 + 2kπ
• x = 6.283 + 2kπ, k  
Solving Trigonometric Equations
Algebraically
Inverse Trigonometric Functions
• Remember, your calculator must be in RADIAN
mode.
• cos x = 0.6
– We can use inverse trig functions to solve for x.
Check the solution graphically
cos x  0 . 6
x  cos
1
(0 . 6 )
x  . 927
x  . 927  2 k π , k  
Why are there two solutions?
x  . 927  2 k π
x  5 . 356  2 k π , k  
Let’s consider the
Unit Circle
Where is x
(cosine) positive?
“All Students Take Calculus”
S
A
sine is positive
all ratios are
positive
cosecant is
positive
T
tangent is
positive
cotangent is
positive
C
cosine is positive
secant is positive
How do we find the other solutions
algebraically?
For Cosine
For Sine
Calculator Solution
Calculator Solution
– Calculator Solution
π – Calculator Solution
cos x = 0.6
cos x  0 . 6
x  cos
1
(0 . 6 )
x  . 927
x  . 927  2 k π , k  
x  5 . 356  2 k π
Your Turn:
• Solve for all solutions algebraically:
cos x = – 0.3
sin x = –0.75
Your Turn:
• Solve for all solutions algebraically:
sin x = 0.5
What about tangent?
• The solution that you get in the calculator is
the only one!
tan x = –5
Your Turn:
• Solve for all solutions algebraically:
1. cos x = –0.2
2. sin x = – ⅓
3. tan x = 3
4. sin x = 4
What’s going on with #4?
• sin x = 4
How would you solve for x if…
3x2 – x = 2
So what if we have…
3 sin2x – sin x = 2
What about…
tan x cos2x – tan x = 0
Your Turn:
• Solve for all solutions algebraically:
5. 4 sin2x = 5 sin x – 1
6. cos x sin2x = cos x
7. sin x tan x = sin x
8. 5 cos2x + 6 cos x = 8
Warm-up (4 m)
1. Solve for all solutions algebraically:
3 sin2x + 2 sin x = 5
2. Explain why we would reject the solution
cos x = 10
3 sin2x + 2 sin x = 5
Explain why we would reject the
solution cos x = 10
What happens if you can’t factor the
equation?
• x2 + 5x + 3 = 0
Quadratic Formula
x
b 
2
b  4 ac
2a
The plus or minus
symbol means that you
actually have TWO
equations!
x2 + 5x + 3 = 0
ax2 + bx + c = 0
Using the Quadratic Equation to Solve
Trigonometric Equations
• You can’t mix trigonometric functions. (Only
one trigonometric function at a time!)
• Must still follow the same basic format:
• ax2 + bx + c = 0
• 2 cos2x + 6 cos x – 4 = 0
• 7 tan2x + 10 = 0
tan2x + 5 tan x + 3 = 0
3 sin2x – 8 sin x = –3
Your Turn:
• Solve for all solutions algebraically:
1. sin2x + 2 sin x – 2 = 0
2. tan2x – 2 tan x = 2
3. cos2x = –5 cos x + 1
Warm-up (4 m)
• Solve for all solutions algebraically:
1. tan x cos x + 3 tan x = 0
2. 2 cos2x + 7 cos x – 1 = 0
tan x cos x + 3 tan x = 0
2 cos2x + 7 cos x – 1 = 0
Seek and Solve!
You have 30 m to complete the seek
and solve. Show all your work on a
sheet of paper because I’m collecting
it for a classwork grade.
Remember me?
sec x 
1
cos x
csc x 
2
1
cot x 
sin x
2
sin x  cos x  1
2
2
1  cot x  csc x
2
2
tan x  1  sec x
1
tan x
Using Reciprocal Identities to Solve
Trigonometric Equations
• Our calculators don’t have reciprocal function
(sec x, csc x, cot x) keys.
• We can use the reciprocal identities to rewrite
secant, cosecant, and cotangent in terms of
cosine, sine, and tangent!
csc x = 2
csc x = ½
cot x cos x = cos x
Your Turn:
• Use the reciprocal identities to solve for
solutions algebraically:
1. cot x = –10
2. tan x sec x + 3 tan x = 0
3. cos x csc x = 2 cos x
Using Pythagorean Identities to Solve
Trigonometric Equations
• You can use a Pythagorean identity to solve a
trigonometric equation when:
– One of the trig functions is squared
– You can’t factor out a GCF
– Using a Pythagorean identity helps you rewrite
the squared trig function in terms of the other trig
function in the equation
2
cos x
–
2
sin x
+ sin x = 0
2
sec x
–2
2
tan x
=0
2
sec x
+ tan x = 3
Your Turn:
• Use Pythagorean identities to solve for all
solutions algebraically:
1. –10 cos2x – 3 sin x + 9 = 0
2. –6 sin2x + cos x + 5 = 0
3. sec2x + 5 tan x = –2