How do you solve $cos (2 θ) + cos (θ) = 0$ $cos(2theta)+cos(theta)=0$ ?

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Answer 1 · 2015-06-08T05:05:31

Use trig identity: $cos 2 a = 2 {cos}^{2} a - 1 .$ $cos 2a = 2cos^2 a - 1.$

$2 {cos}^{2} x + cos x - 1 = 0 .$ $2cos^2 x + cos x - 1 = 0.$

Call cos x = t, solve the quadratic equation: y = 2t^2 + t - 1 = 0

Since a - b + c = 0, use shortcut. One real root is t = -1 and the other is $(- \frac{c}{a} = \frac{1}{2} .)$ $(-c/a = 1/2.)$

a. cos x = t = -1 --> $x = π$ $x = pi$

b. $cos x = t = \frac{1}{2} - \to x = \pm \frac{π}{3}$ $cos x = t = 1/2 --> x = +- pi/3$

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