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Published byMarie Eyer Modified about 1 year ago

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GET OUT PAPER FOR NOTES!!!

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Warm-up (3:30 m) 1.Solve for all solutions graphically: sin 3 x = –cos 2 x 2.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?

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sin 3 x = –cos 2 x

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cos x = 1 x = 0 + 2kπ x = 6.283 + 2kπ,

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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

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Why are there two solutions? Let’s consider the Unit Circle Where 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 CosineFor Sine Calculator Solution – Calculator Solution Calculator Solution π – Calculator Solution

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cos x = 0.6

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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 = –0.22. sin x = – ⅓ 3. tan x = 34. 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… 3x 2 – x = 2

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So what if we have… 3 sin 2 x – sin x = 2

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What about… tan x cos 2 x – tan x = 0

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Your Turn: Solve for all solutions algebraically: 5. 4 sin 2 x = 5 sin x – 1 6. cos x sin 2 x = cos x 7. sin x tan x = sin x8. 5 cos 2 x + 6 cos x = 8

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Warm-up (4 m) 1. Solve for all solutions algebraically: 3 sin 2 x + 2 sin x = 5 2.Explain why we would reject the solution cos x = 10

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3 sin 2 x + 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? x 2 + 5x + 3 = 0 The plus or minus symbol means that you actually have TWO equations! Quadratic Formula

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x 2 + 5x + 3 = 0 ax 2 + 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: ax 2 + bx + c = 0 2 cos 2 x + 6 cos x – 4 = 0 7 tan 2 x + 10 = 0

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tan 2 x + 5 tan x + 3 = 0

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3 sin 2 x – 8 sin x = –3

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Your Turn: Solve for all solutions algebraically: 1.sin 2 x + 2 sin x – 2 = 0 2.tan 2 x – 2 tan x = 2 3.cos 2 x = –5 cos x + 1

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Warm-up (4 m) Solve for all solutions algebraically: 1.tan x cos x + 3 tan x = 0 2.2 cos 2 x + 7 cos x – 1 = 0

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tan x cos x + 3 tan x = 0

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2 cos 2 x + 7 cos x – 1 = 0

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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.

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Remember me?

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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 = 2csc 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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cos 2 x – sin 2 x + sin x = 0

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sec 2 x – 2 tan 2 x = 0

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sec 2 x + tan x = 3

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Your Turn: Use Pythagorean identities to solve for all solutions algebraically: 1.–10 cos 2 x – 3 sin x + 9 = 0 2.–6 sin 2 x + cos x + 5 = 0 3.sec 2 x + 5 tan x = –2

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