trigonometric equations with square roots

[Maryanne]

I get confused on how to solve trigonmetic equations when there's a square root of a number. Any hints would be appreciated.

Solve the following, over the interval 0 < θ < 2pi

(tan θ - 1)(tan θ + square root of 3) = 0

So, Let x= tan θ

(x - 1)(x + square root of three) = 0

x^2 + square root of three(x) - 1x - square root of three = 0

I don't know what to do next.

 

[stapel]

They gave it to you in factored form. Why multiply, then re-factor, and solve? Why not just solve the factors?

Using "@" for "theta", they've given you:

. . . . .tan(@) - 1 = 0

. . . . .tan(@) + sqrt[3] = 0

Then:

. . . . .tan(@) = 1

. . . . .tan(@) = -sqrt[3]

Read the solutions from that table of values they gave you.

Eliz.

 

[Maryanne]

o, wow I was making it way to complicated thanks