Solve Trig Equation

So tanx sin²x = 2 tanx is equivalent to tanx(sin²x − 2) = 0 Then tanx = 0 or sin²x − 2 = 0, now what?

• solutions come only from tanx = 0,

• solutions come only from sin²x − 2 = 0

• solutions come from where sin x is undefined

• solutions come from where tan x is undefined

Tags

CCSS.HSF.TF.B.7

To solve this equation, cos²x + sinx = 1, replace cos²x with

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

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

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

• 0, π

• 0, π, ½π

• ½π

• 0, 90, 270

Tags

CCSS.HSF.TF.B.7

csc²x = 2 is equivalent to sin²x = ½

• True

• False

Tags

CCSS.HSF.TF.C.8

To solve the equation sec²x − secx = 2

• start by subtracting 2 from both sides and then factor

• add sec x to both sides and factor

• factor out secx and set each factor equal to 2

• sec x is never bigger than 1, so there are no solutions

Tags

CCSS.HSF.TF.B.7

• 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 for the unknown variable on the interval 0≤x<2π

sinx + 2sinxcosx = 0

• 2π/3, 4π/3, 0, π

• 2π/3, 4π/3

• 0, π

• π/3, 5π/3

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
cos² x + sin x + 1 = 0

• π

• 3π/2

• π/6, 5π/6, 3π/2

• No solution

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