Quick Check: Solving Trig Equations

Which of the following is NOT
a solution to
sin θ = √(3) / 2 ?

• π / 3
• 2π / 3
• 5π / 3
• 7π / 3

Tags

CCSS.HSF.TF.B.7

Solve
 $\cos^2\theta=\frac{1}{2}$  
on the interval [0, 2π). HInt: When square rooting don't forget  $\pm$  .  May have to rationalize denominator.

• θ = π /4, 7π /4

• θ = π /4, 3π /4

• θ = 3π /4, 5π /4

• θ = π /4, 3π /4, 5π /4, 7π /4

 $\pm$  Solve on the domain [0, 2π)
 $4\sin^2x=3$   HInt: $\pm$  

• π/6, 11π/6

• π/3, 2π/3

• π/6, 5π/6, 7π/6, 11π/6

• π/3, 2π/3, 4π/3, 5π/3

Solve on the domain [0, 2π)

2sin x cos x = √2 cos x. HInt: move everything to the left, factor out cosx and then set each factor equal to zero.

• π/2, 3π/2

• 0, π/4, 3π/4

• π/2, 3π/2, π/4, 3π/4

• No solution

Solve on the domain [0, 2π)
 $\cos^2x+\sin x+1$   = 0. Hint: Change $\cos^2x\ to\ 1-\sin^2x$  

• π

• 3π/2

• π/6, 5π/6, 3π/2

• No 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

Why doesn't 2cosx − 3 = 0 have solutions?

• cos x is never bigger than one

• cos x is never equal to a fraction

• Actually, this equation does have a solution, x = π

• This equation will have a solution tomorrow.

Tags

CCSS.HSF.TF.B.7

To solve this equation, cos²x + sinx = 1, replace cos²x with

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

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

Tags

CCSS.HSF.TF.B.7