Question

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

Answer 1

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

Explanation:

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]