Solving Trig Identities

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?

• cosx is never bigger than one

• cosx 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

Choose a good way to start solving this equation

tanx sin²x = 2 tanx

• divide tanx from both sides

• factor out tanx

• subtract 2 tanx from the left and then factor tanx out

• cancel sin²x out

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Factor 0 = sin²x − sinx

• 0 = sinx(sinx)

• 0 = sinx(1 − sinx)

• 0 = cosx(sinx − 1)

• 0 = sinx(sinx − 1)

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

On the domain [0, 2π), solve this equation 0 = sinx(sinx − 1)

• 0, π

• 0, π, ½π

• ½π

• 0, 90, 270

Tags

CCSS.HSA.REI.B.4

CCSS.HSF.TF.A.1

CCSS.HSF.TF.A.2

CCSS.HSF.TF.A.4

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.A.1

CCSS.HSA.APR.B.2

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Solve the following for cos²θ:
sin²θ + cos²θ = 1

• cos²θ = 1 - sin²θ
• cosθ = 1 + sin²θ
• cosθ = 1 - sinθ
• cos²θ = Adj/hyp

Tags

CCSS.HSA.CED.A.4

CCSS.HSF.TF.C.8

Solve the following for tan²θ:
1 + tan²θ = sec²θ 

• tan²θ = sec²θ + 1
• tan²θ = sec²θ - 1
• tanθ = secθ - 1
• tan²θ = opp/adj

Tags

CCSS.HSA.REI.B.4

CCSS.HSF.TF.C.8

Rewrite tan(x) in terms of sin(x) and cos(x).

• tan(x) = cos(x)/sin(x)
• tan(x) = 1/cot(x)
• tan(x) = sin(x)/cos(x)
• tan(x) = opp/adj

Tags

CCSS.HSF.TF.C.8