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

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

1 Answer
Nov 5, 2017

Solution: In Interval: 0 <= theta<=2pi, theta=90^0, theta=330^00θ2π,θ=900,θ=3300

Explanation:

Interval: 0 <= theta<=2pi0θ2π

cos 2theta + sin theta=0 or 1-2sin^2 theta+sin theta=0cos2θ+sinθ=0or12sin2θ+sinθ=0.
As [cos 2theta= 1-2sin^2 theta][cos2θ=12sin2θ]

2sin^2 theta-sin theta-1 =02sin2θsinθ1=0 or

2sin^2 theta-2sin theta +sin theta -1 =02sin2θ2sinθ+sinθ1=0 or

2sin theta(sin theta- 1) +1(sin theta -1) =02sinθ(sinθ1)+1(sinθ1)=0 or

(sin theta- 1)(2sin theta+1)=0(sinθ1)(2sinθ+1)=0. Either sin theta- 1=0sinθ1=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]