assessment ch 5a part 2 no calculator allowed form b name: billc period: 1 score: 10.5/14 - 3.5 standard 5a…

assessment ch 5a part 2 no calculator allowed form b name: billc period: 1 score: 10.5/14 - 3.5 standard 5a - s2: using trigonometric identities evaluate and solve equation in various forms. find all solutions in the interval 0, 2π). all work must be shown. (2 points each) 1) cos 2x + cos x = 0 1) - 2 2) π

assessment ch 5a part 2 no calculator allowed form b name: billc period: 1 score: 10.5/14 - 3.5 standard 5a - s2: using trigonometric identities evaluate and solve equation in various forms. find all solutions in the interval 0, 2π). all work must be shown. (2 points each) 1) cos 2x + cos x = 0 1) - 2 2) π

Answer

Explanation:

Step1: Use double - angle identity

Recall $\cos2x = 2\cos^{2}x - 1$. The equation $\cos2x+\cos x = 0$ becomes $2\cos^{2}x - 1+\cos x = 0$. Let $t=\cos x$, then the equation is $2t^{2}+t - 1 = 0$.

Step2: Solve the quadratic equation

Factor the quadratic equation $2t^{2}+t - 1=(2t - 1)(t + 1)=0$. So $2t-1 = 0$ or $t + 1=0$. Solving $2t-1 = 0$ gives $t=\frac{1}{2}$, and solving $t + 1=0$ gives $t=-1$.

Step3: Substitute back $\cos x$ for $t$

When $\cos x=\frac{1}{2}$, in the interval $[0,2\pi)$, $x=\frac{\pi}{3}$ or $x = \frac{5\pi}{3}$. When $\cos x=-1$, in the interval $[0,2\pi)$, $x=\pi$.

Answer:

$x=\frac{\pi}{3},\pi,\frac{5\pi}{3}$