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Get out paper for notes!!!. Warm-up (3:30 m). Solve for all solutions graphically: sin 3 x = –cos 2 x Molly found that the solutions to cos x = 1 are x = 0 + 2kπ AND x = 6.283 + 2kπ, . Is Molly’s solution correct? Why or why not?. sin 3 x = –cos 2 x. cos x = 1. x = 0 + 2kπ - PowerPoint PPT PresentationTRANSCRIPT
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GET OUT PAPER FOR NOTES!!!
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Warm-up (3:30 m)
1. Solve for all solutions graphically: sin3x = –cos2x2. Molly found that the solutions to cos x = 1 are
x = 0 + 2kπ AND x = 6.283 + 2kπ, . Is Molly’s solution correct? Why or why not?
k
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sin3x = –cos2x
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cos x = 1
• x = 0 + 2kπ• x = 6.283 + 2kπ, k
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Solving Trigonometric Equations Algebraically
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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.
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Check the solution graphically
k,πk2927.x927.x
)6.0(cosx
6.0xcos1
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Why are there two solutions?
k,πk2356.5xπk2927.x
Let’s consider the Unit CircleWhere is x
(cosine) positive?
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“All Students Take Calculus”AS
CT
all ratios are positive
sine is positive
tangent is positive
cosine is positive
cosecant is positive
cotangent is positive
secant is positive
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How do we find the other solutions algebraically?
For Cosine For Sine
Calculator Solution
– Calculator Solution
Calculator Solution
π – Calculator Solution
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cos x = 0.6
k,πk2927.x927.x
)6.0(cosx
6.0xcos1
πk2356.5x
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Your Turn:
• Solve for all solutions algebraically:cos x = – 0.3
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sin x = –0.75
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Your Turn:
• Solve for all solutions algebraically:sin x = 0.5
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What about tangent?
• The solution that you get in the calculator is the only one!
tan x = –5
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Your Turn:
• Solve for all solutions algebraically:1. cos x = –0.2 2. sin x = – ⅓
3. tan x = 3 4. sin x = 4
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What’s going on with #4?
• sin x = 4
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How would you solve for x if…
3x2 – x = 2
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So what if we have…
3 sin2x – sin x = 2
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What about…
tan x cos2x – tan x = 0
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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
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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
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3 sin2x + 2 sin x = 5
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Explain why we would reject the solution cos x = 10
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What happens if you can’t factor the equation?
• x2 + 5x + 3 = 0
a2ac4bbx
2
The plus or minus symbol means that you
actually have TWO equations!
Quadratic Formula
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x2 + 5x + 3 = 0ax2 + bx + c = 0
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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
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tan2x + 5 tan x + 3 = 0
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3 sin2x – 8 sin x = –3
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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
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Seek and Solve!
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Remember me?
xtan1xcot
xsin1xcsc
xcos1xsec
xsec1xtan
xcscxcot1
1xcosxsin
22
22
22
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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!
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csc x = 2 csc x = ½
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cot x cos x = cos x
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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
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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
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cos2x – sin2x + sin x = 0
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sec2x – 2 tan2x = 0
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sec2x + tan x = 3
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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