How do you solve #sin 2theta - cos theta = 0 # between 0 and 2pi?

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Answer 1 · 2016-02-15T08:43:05

in Degrees:#" " " "theta=30^@, 90^@, 150^@, 270^@#
in Radians:#" " " "theta=pi/6, pi/2, (5pi)/6, (3pi)/2#

The given: Solve
#sin 2theta-cos theta=0#

because
#sin 2theta=2*sin theta cos theta#
it follows

#2 sin theta cos theta-cos theta=0#
factor out #cos theta#

#cos theta(2 sin theta-1)=0#
Equate both factors to zero

#cos theta=0# and #2 sin theta-1=0#
#theta=cos^-1 (0)=90^@, 270^@#

also

#2 sin theta = 1#
#sin theta = 1/2#
#theta=sin^-1 (1/2)=30^@,150^@#

the solutions are:

#theta=30^@, 90^@, 150^@, 270^@#

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