# solving trig equations

• Feb 7th 2007, 11:17 AM
Tom G
solving trig equations
Can someone please check that I've got the correct answers as they don't appear to fit.

Q: cos2x = -(√3)/2 -180<=x<=180

I've got 2x=210, 330, 570, 690
x=105, 165, 285, 345

We're currently learning about using CAST diagrams to solve trigonometric equations, so if any anyone can explain in terms of CAST diagrams I'd be grateful.
• Feb 7th 2007, 12:33 PM
Soroban
Hello, Tom!

You picked some incorrect angles . . .

Quote:

cos(2x) = -(√3)/2, . -180° < x < 180°

Cosine is negative in Quadrants 2 and 3.

. . 2x .= .-210°, -150°, 150°, 210°

Therefore: .x .= .-105°, -75°, 75°, 105°

• Feb 7th 2007, 10:42 PM
CaptainBlack
Quote:

Originally Posted by Tom G
Can someone please check that I've got the correct answers as they don't appear to fit.

Q: cos2x = -(√3)/2 -180<=x<=180

I've got 2x=210, 330, 570, 690
x=105, 165, 285, 345

We're currently learning about using CAST diagrams to solve trigonometric equations, so if any anyone can explain in terms of CAST diagrams I'd be grateful.

Your answers other than the first are not in the specified range for x.

My calculations give: x=75, 105, -75, -105 degrees. Which check out
numericaly.

RonL