Asked by Autumn
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!
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!
Answers
Answered by
Steve
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.
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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