Solving Trig Equations with Basic Factoring

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

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Which is a correct way to solve the equation

cosx(2cosx − 3) = 0?

• divide cos x from both sides

• set each factor equal to 0 and solve

• distribute cosx into the parentheses

• guess and hope for the best

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.

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

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

The square root of 2 is bigger than 1.

• true

• false

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

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

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

Factor 0 = sin²x − sinx

• 0 = sinx(sinx)

• 0 = sinx(1 − sinx)

• 0 = cosx(sinx − 1)

• 0 = sinx(sinx − 1)

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

• 0, π

• 0, π, ½π

• ½π

• 0, 90, 270