Solving Trig Equations

Solve

cos²θ = ½

on the interval from [0, 2π)

• θ = π /4, 7π /4

• θ = π /4, 3π /4

• θ = 3π /4, 5π /4

• θ = π /4, 3π /4, 5π /4, 7π /4

Solve on the interval from

 $\left[0,\ 2\pi\right)$ 

• Undefined

• π/3, 5π/3

• π/6, 11π/6

• 2π/3, 4π/3

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)

1 - sin²x + sin x + 1 = 0

• π

• 3π/2

• π/6, 5π/6, 3π/2

• No solution

Solve on the domain [0, 2π)

cos x + 2 = 3 cos x

• 0, 2π/3, 4π/3

• π

• 0

• Undefined

Solve for x on the interval from

 $\left[0,\ 2\pi\right)$ 

• π

• π/3

• π/2

• -π

How many solutions would this equation have on the interval from  $\left[0,\ 2\pi\right)$ ?

2cos(2x)= -√2

• 4 solutions

• 2 solutions

• 8 solutions

• 1 solution

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

Organize this equation so it can be factored:

1−sin²x + sinx = 1

• move all the terms to one side: 0 = sin²x − sinx

• cancel the ones and cancel a sinx

• replace sin²x with 1 − cos²x

• use this: 0 = sin²x − sinx + 2