Factoring Trig Equations

Factor Completely:
9cos²x - 12cosx +4

• (3cosx + 4) ( 3cosx +1)

• (3cosx - 2)²

• (cosx -2)(cosx - 2)

• (9cosx -2) (cosx -2)

Factor Completely:
1 - 4sin²x

• (1 - 2sinx)²

• (sinx -2) (sinx +2)

• (2sinx -1) (2sinx +1)

• (1 - 2sinx) (1 +2sinx)

Factor Completely:
cos²y - cosy - 6

• (cosy - 3) (cosy +2)

• (cos y - 3)²

• (cosy +3) (cosy -2)

• (cosy -6) (cosy +1)

Factor Completely:
9csc²x - 12cscx +4

• (cscx - 2) (cscx -2)

• (3cscx -2) (3cscx + 2)

• (3cscx -2)²

• (3 cscx + 2)²

Factor 0 = sin²x − sinx

• 0 = sinx(sinx)

• 0 = sinx(1 − sinx)

• 0 = cosx(sinx − 1)

• 0 = sinx(sinx − 1)

Factor Completely: 2cos² x+ 5cos x -3

• 2(cos x -1)(cos x +3)

• (cos x +3)(cos x-1)

• (2cos x +3) (cos x -1)

• (2cos x - 1)(cos x +3)

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

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Which of these shows the first step in solving 2cos²x − 3cosx = 0 ?

• 2cos²x = 3cosx

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• (cosx+3)(cosx-1) = 0

Solve for 0≤x≤2π

2cos²x + cosx = 0

• π/2, 3π/2

• 0, π,2π

• 2π/3, 4π/3

• π/3, 5π/3

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