# Math

Solve the equation 2cos2x = √3 for 0°≤x≤360°

I did this:
cos2x = √3 /2
2x=30
x=15
x=15, 165, 195, 345

Is this correct?

Solve the equation √3 sin2x + cos2x = 0 for -π≤x≤π

No idea how to approach this one

Thanks a bunch for the help!

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1. 2cos2x = √3
cos2x = √3/2
2x = 30°,330° + 360n
x = 14,165 + 180n
x = 15°,195°,165°,345°
You are correct

√3 sin2x + cos2x = 0
√3 sin2x = -cos2x
tan2x = -1/√3
tan < 0 in QII,QIV
and so on as in the other problem.

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posted by Steve

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