solve-at-0-pi-cos-cos-2-cos-3-0-

Question Number 33116 by abdo imad last updated on 10/Apr/18

s o l v e a t [0, π] c o s α + c o s (2 α) + c o s (3 α) = 0

Answered by MJS last updated on 10/Apr/18

\cos 2 α = {2 \cos}^{2} α - 1

\cos 3 α = {4 \cos}^{3} α - 3 \cos α

\cos α + {2 \cos}^{2} α - 1 + {4 \cos}^{3} α - 3 \cos α = 0

{4 \cos}^{3} α + {2 \cos}^{2} α - 2 \cos α - 1 = 0

\cos^{3} α + \frac{1}{2} \cos^{2} α - \frac{1}{2} \cos α - \frac{1}{4} = 0

trying \cos α = \pm \frac{1}{4}; \pm \frac{1}{2} \Rightarrow

\Rightarrow {(\cos α)}_{1} = - \frac{1}{2} \Rightarrow α_{1} = \frac{2 π}{3}

\frac{\cos^{3} α + \frac{1}{2} \cos^{2} α - \frac{1}{2} \cos α - \frac{1}{4}}{\cos α + \frac{1}{2}} = 0

\cos^{2} α - \frac{1}{2} = 0

{(\cos α)}_{2} = - \frac{\sqrt{2}}{2} \Rightarrow α_{2} = \frac{3 π}{4}

{(\cos α)}_{3} = \frac{\sqrt{2}}{2} \Rightarrow α_{3} = \frac{π}{4}

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