Solve Trigonometric Equations

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

• π/2; 3π/2; π/4; 3π/4

• π/2; 3π/2

• π/4; 3π/4

• π/4

Tags

CCSS.HSF.TF.B.7

Find the general solution of the equation.

• 0

• 2π

• 2πn

• No Solution

Tags

CCSS.HSF.TF.B.7

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

• 0, π, π/6, 5π/6, 7π/6, 11π/6

• 0, π/6, 7π/6

• π, 5π/6, 11π/6

• π, π/6, 5π/6, 7π/6, 11π/6

Tags

CCSS.HSF.TF.B.7

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

• π/6; 5π/6

• π/6; 5π/6; 7π/6;11π/6

• 7π/6;11π/6

• π/6; 7π/6

Tags

CCSS.HSF.TF.B.7

So tanx sin²x = 2 tanx is equivalent to tanx(sin²x − 2) = 0

Then tanx = 0 or sin²x − 2 = 0, now what?

• solutions come only from tanx = 0,

• solutions come only from sin²x − 2 = 0

• solutions come from where sin x is undefined

• solutions come from where tan x is undefined

Tags

CCSS.HSF.TF.B.7

Solve the equation (secx − 2)(secx + 1) = 0 on [0, 360⁰)

• 60⁰, 180⁰, 300⁰

• 30⁰, 180⁰,330⁰

• 60⁰, 120⁰

• 180⁰

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

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

Tags

CCSS.HSF.TF.B.7

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

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x