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

Question

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

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Camila 2 weeks 2021-09-07T15:55:33+00:00 2 Answers 0

Answers ( )

    0
    2021-09-07T15:57:16+00:00

    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)

    0
    2021-09-07T15:57:21+00:00

    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

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