Solving Trig Equations

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

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)
cos x + 2 = 3 cos x

• 0, 2π/3, 4π/3

• π

• 0

• No solution

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0,2π)
1/4sin x + 1= 0

• 0

• 7π/4

• 5π/4

• No solution

Tags

CCSS.HSF.TF.B.7

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

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Tags

CCSS.HSF.TF.B.7

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

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

csc²x = 2 is equivalent to sin²x = ½

• True

• False

Tags

CCSS.HSF.TF.C.8

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