Solving Trigonometric Equations Algebraically

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

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

• cosx is never bigger than one

• cosx is never equal to a fraction

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

• This equation will have a solution tomorrow.

Solve for 0≤x≤2π

2cos²x = cosx + 1

• 2π/3, 4π/3

• π/3, 2π/3

• 0, 2π

• π

Solve for 0≤x≤2π

2cos²x + 3cosx = -1

• 0, 2π

• π

• 2π/3, 4π/3

• π/3, 5π/3

Solve for 0≤x≤2π

6sin²x - sinx - 1 = 0

• 3.48, 5.94

• 0.33, 2.81

Solve [0,2π] 4cos²x - 7cosx - 2 = 0

• 1.82, 4.46

• 1.23, 3.87

• 2.07, 5.73

• no solution

Solve for 0≤x≤2π

tanx = 1

• π/2, π

• π/4, 3π/4

• π/4, 5π/4

• 3π/4, 7π/4

Solve for 0≤x≤2π

tan²x - 9 = 0

• 1.25, 4.39

• 0.78, 3.27

• 1.89, 5.03

• π/3, 5π/3

7 + 2cosx = 8sin²x

Solve [0, 2π]