How do you find the exact value of #sin2theta-costheta=0# in the interval #0<=theta<360#?

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How do you find the exact value of #sin2theta-costheta=0# in the interval #0<=theta<360#?

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Answer 1 · 2016-12-19T05:33:19

#pi/6; pi/2; (3pi)/2; (5pi)/6#

Explanation:

Replace sin 2t by 2sint.cos t -->
2sin t.cos t - cos t = 0
cos t(2sin t - 1) = 0
Use trig table of special arcs and unit circle -->
a. cos t = 0 --> arc #t = pi/2, and t = (3pi)/2#
b. (2sin t - 1) = 0 --> sin t = 1/2 -->
#t = pi/6 and t = (5pi)/6#
Answers for #(0, 2pi)#:
#pi/6, pi/2, (3pi)/2, (5pi)/6#

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How do you find the exact value of #sin2theta-costheta=0# in the interval #0<=theta<360#?

1 Answer

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