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

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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

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Answer 1 · 2017-11-05T11:03:42

Solution: In Interval: $0 \leq θ \leq 2 π, θ = 90^{0}, θ = 330^{0}$ $0 <= theta<=2pi, theta=90^0, theta=330^0$

Explanation:

Interval: $0 \leq θ \leq 2 π$ $0 <= theta<=2pi$

$cos 2 θ + sin θ = 0 or 1 - 2 {sin}^{2} θ + sin θ = 0$ $cos 2theta + sin theta=0 or 1-2sin^2 theta+sin theta=0$ .
As $[cos 2 θ = 1 - 2 {sin}^{2} θ]$ $[cos 2theta= 1-2sin^2 theta]$

$2 {sin}^{2} θ - sin θ - 1 = 0$ $2sin^2 theta-sin theta-1 =0$ or

$2 {sin}^{2} θ - 2 sin θ + sin θ - 1 = 0$ $2sin^2 theta-2sin theta +sin theta -1 =0$ or

$2 sin θ (sin θ - 1) + 1 (sin θ - 1) = 0$ $2sin theta(sin theta- 1) +1(sin theta -1) =0$ or

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

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]

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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

Explanation:

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