Solve the following equation exactly on the interval 0 ≤ θ ≤ 2π ? cos 2θ + sin θ = 0

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I'm really confused on how to use trigonometric functions to find the answer

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Answer 1 · 2017-11-05T11:03:42

Solution: In Interval: #0 <= theta<=2pi, theta=90^0, theta=330^0#

Interval: #0 <= theta<=2pi#

#cos 2theta + sin theta=0 or 1-2sin^2 theta+sin theta=0#.
As #[cos 2theta= 1-2sin^2 theta]#

#2sin^2 theta-sin theta-1 =0# or

#2sin^2 theta-2sin theta +sin theta -1 =0# or

#2sin theta(sin theta- 1) +1(sin theta -1) =0# or

#(sin theta- 1)(2sin theta+1)=0#. Either #sin theta- 1=0#

or #2sin theta+1=0 :. sin theta=1 or 2sin theta=-1#

When #sin theta =1 ; sin(pi/2)=1 :. theta=pi/2#

When #2sin theta = -1 or sin theta= -1/2#

#sin (-pi/6)=-1 :. theta= -pi/6 or theta=(2pi-pi/6)=(11pi)/6#

Solution: In Interval: #0 <= theta<=2pi, theta=pi/2, (11pi)/6#

In degree mode: Solution: In Interval: #0 <= theta<=2pi, #

#theta=90^0, theta=330^0#[Ans]