How do you find the exact value of #3sin2theta=tantheta# in the interval #0<=theta<2pi#?

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Answer 1 · 2017-01-25T17:45:22

#3(2sinthetacostheta) = sintheta/costheta#

#6sinthetacostheta = sintheta/costheta#

#6sinthetacos^2theta = sintheta#

#6cos^2theta = 1#

#cos^2theta = 1/6#

#costheta = +- 1/sqrt(6)#

#theta = arccos(1/sqrt(6)), pi - arccos(1/sqrt(6)), pi + arccos(1/sqrt(6)), 2pi - arccos(1/sqrt(6))#

Hopefully this helps!