Solving Trig Equations II

Solve
cos²θ = ½ 
on θ∈[0, 2π)

• θ = π /4, 7π /4
• θ = π /4, 3π /4
• θ = 3π /4, 5π /4
• θ = π /4, 3π /4, 5π /4, 7π /4

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
1/2sec x - 1 = 0

• π/6, 5π/6
• π/3, 5π/3
• 2π/3, 4π/3
• 7π/6, 11π/6

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
4sin²x = 3

• π/6, 11π/6
• π/3, 2π/3
• π/6, 5π/6, 7π/6, 11π/6
• π/3, 2π/3, 4π/3, 5π/3

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
2sin x cos x = √2 cos x

• π/2, 3π/2
• 0, π/4, 3π/4
• π/2, 3π/2, π/4, 3π/4
• No solution

Tags

CCSS.HSF.TF.B.7

Which of these is equivalent to 2cos²x − 3cosx = 0 ?

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Tags

CCSS.HSF.TF.B.7

Why doesn't 2cosx − 3 = 0 have solutions?

• cos x is never bigger than one

• cos x is never equal to a fraction

• Actually, this equation does have a solution, x = π

• This equation will have a solution tomorrow.

Tags

CCSS.HSF.TF.B.7

Organize this equation so it can be factored:

1−sin²x + sinx = 1

• move all to the right: 0 = sin²x − sinx

• cancel the ones and cancel a sinx

• replace sin²x with 1 − cos²x

• use this: 0 = sin²x − sinx + 2

Tags

CCSS.HSF.TF.B.7

Factor 0 = sin²x − sinx

• 0 = sinx(sinx)

• 0 = sinx(1 − sinx)

• 0 = cosx(sinx − 1)

• 0 = sinx(sinx − 1)

Tags

CCSS.HSF.TF.B.7

csc²x = 2 is equivalent to sin²x = ½

• True

• False

Tags

CCSS.HSF.TF.C.8

Factor: sec²x − secx − 2

• (sec x)(secx − 2)

• (secx − 2)(secx − 1)

• (secx − 2)(secx + 1)

• (secx + 2)(secx − 1)

Tags

CCSS.HSA.APR.C.4