Deriving Trig Equations

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

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

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

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

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

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

Factor 0 = sin²x − sinx

• 0 = sinx(sinx)

• 0 = sinx(1 − sinx)

• 0 = cosx(sinx − 1)

• 0 = sinx(sinx − 1)

Find the general solution of the equation.

• π/4+2πpn

• π/4+2πn; 7π/4+2πpn

• 7π/4+2πpn

• π/4, 7π/4

Find the general solution of the equation.

• 0

• 2π

• 2πn

• No Solution

Find the general solution of the equation.

• π/6+2πn; 11π/6+2πn

• π/6+2πn

• 11π/6+2πn

• π/6; 11π/6

Find the general solution of the equation.

• π/6+2πn

• 5π/6+2πn

• π/6; 5π/6

• π/6+2πn; 5π/6+2πn

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

• π/6; 5π/6; 7π/6; 11π/6

• π/6; 5π/6

• 7π/6; 11π/6

• π/6+2πn; 5π/6+2πn