How do you solve $cos (2 \cdot θ) + cos (θ) = 0$ $Cos(2*theta)+cos(theta)=0$ ?

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Answer 1 · 2016-11-19T02:44:16

Use the identity ${cos}^{2} β = 1 - 2 {sin}^{2} β$ $cos^2beta = 1- 2sin^2beta$ .

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

$1 - 2 (1 - {cos}^{2} θ) + cos θ = 0$ $1 - 2(1 - cos^2theta) + costheta = 0$

$1 - 2 + 2 {cos}^{2} θ + cos θ = 0$ $1 - 2 + 2cos^2theta + costheta = 0$

$2 {cos}^{2} θ + cos θ - 1 = 0$ $2cos^2theta + costheta - 1 = 0$

$2 {cos}^{2} θ + 2 cos θ - cos θ - 1 = 0$ $2cos^2theta + 2costheta - costheta - 1 = 0$

$2 cos θ (cos θ + 1) - (cos θ + 1) = 0$ $2costheta(costheta + 1) - (costheta + 1) = 0$

$(2 cos θ - 1) (cos θ + 1) = 0$ $(2costheta - 1)(costheta + 1) = 0$

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, π + 2 π n$ $theta = pi/3 + 2pin, (2pi)/3 + 2pin, pi + 2pin$ .

Hopefully this helps!

How do you solve Cos(2*theta)+cos(theta)=0cos(2⋅θ)+cos(θ)=0Cos(2*theta)+cos(theta)=0?