Solving Trig Equations with Factoring

• A
• B
• C
• D

Find the solution over the interval [0, 2π)

• 0, π, π/6, 5π/6,

• 0, π/6, 7π/6

• π, 5π/6, 11π/6

• π, π/6, 5π/6, 7π/6, 11π/6

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

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

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

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)

Factor: sec²x − secx − 2

• (sec x)(secx − 2)

• (secx − 2)(secx − 1)

• (secx − 2)(secx + 1)

• (secx + 2)(secx − 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)

Solve for 0≤x≤2π

3sin²x + sinx - 4 = 0

• 0, π

• π/2

• π/6, 5π/6

• 3π/2

Solve for 0≤x≤2π

2cos²x + cosx = 0

• π/2, 3π/2

• 0, π,2π

• 2π/3, 4π/3

• π/3, 5π/3