Question

Solve the following equation.6cos²x - 7cosx + 1 = 0, for 0° ≤ x ≤ 360°​

Answer 1

Answer:

For 0° ≤ x ≤ 360°​ we have:

x=0, 360, arccos(1/6), -arccos(1/6)

Step-by-step explanation:

Replace cos(x) by y

6y²-7y+1=0

6y²-6y-y+1=0

6y(y-1) -(y-1)=0

(y-1)(6y-1)=0

Roots:

y=1, cos(x)=1, x=0 and 360

y=1/6, cos(x)=1/6, x=arccos(1/6) and x=-arccos(1/6)

Answer 2

Answer:

x = 0° or x = 80° or x = 260° or x = 360°

Step-by-step explanation:

$6\cos^2x-7\cos x+1=0\qquad\text{substitute}\ \cos x=t,\ -1\leq t\leq 1\\\\6t^2-7t+1=0\\\\6t^2-6t-t+1=0\\\\6t(t-1)-1(t-1)=0\\\\(t-1)(6t-1)=0\iff t-1=0\ \vee\ 6t-1=0\\\\t-1=0\qquad\text{add 1 to both sides}\\t=1\\\\6t-1=0\qquad\text{add 1 to both sides}\\6t=1\qquad\text{divide both sides by 6}\\t=\dfrac{1}{6}$

$t=1\to\cos x=1\Rightarrow x=0^o\ \vee\ x=360^o\\\\t=\dfrac{1}{6}=0.1666...\to\cos x=0.1666...\to x\approx80^o\ \vee\ x\approx260^o$