Solving Trig Equations with Basic Factoring

• A
• B
• C
• D

Tags

CCSS.HSF.TF.B.7

• A
• B
• C
• D

Tags

CCSS.HSF.TF.B.7

Solve for x

don't forget cot = cos/sin

• π

• π/3

• π/2

• -π

Tags

CCSS.HSF.TF.B.7

Solve: 4sinθ + 1 = 3 State all solutions in the interval [0, 2π)

• π/ 6, 7π/6

• π/6, 5π/ 6

• 5π/6, 7π/6

• 7π/6, 11π/6

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.HSA.REI.C.7

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)

Tags

CCSS.HSA.APR.C.4

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

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

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

Tags

CCSS.HSF.TF.B.7

Solve for 0≤x≤2π

2cos²x + cosx = 0

• π/2, 3π/2

• 0, π,2π

• 2π/3, 4π/3

• π/3, 5π/3

Tags

CCSS.HSF.TF.B.7