How do you solve $2 = 4 {cos}^{2} x + 1$ $2=4cos^2x+1$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve $2 = 4 {cos}^{2} x + 1$ $2=4cos^2x+1$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-12-20T01:11:04

$x \in {\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}}$ $x in {pi/3, (2pi)/3, (4pi)/3, (5pi)/3}$

Explanation:

$2 = 4 {cos}^{2} (x) + 1$ $2=4cos^2(x)+1$

$\Rightarrow 4 {cos}^{2} (x) - 1 = 0$ $=> 4cos^2(x)-1 = 0$

$\Rightarrow {cos}^{2} (x) - \frac{1}{4} = 0$ $=> cos^2(x)-1/4 = 0$

$\Rightarrow (cos (x) + \frac{1}{2}) (cos (x) - \frac{1}{2}) = 0$ $=> (cos(x)+1/2)(cos(x)-1/2) = 0$

$\Rightarrow cos (x) + \frac{1}{2} = 0 or cos (x) - \frac{1}{2} = 0$ $=> cos(x)+1/2 = 0 or cos(x)-1/2 = 0$

$\Rightarrow cos (x) = - \frac{1}{2} or cos (x) = \frac{1}{2}$ $=> cos(x) = -1/2 or cos(x) = 1/2$

Checking a unit circle or relying on knowledge of common angles, we find that with the restriction $x \in [0, 2 π]$ $x in [0, 2pi]$ , we have

$cos (x) = - \frac{1}{2} \Leftrightarrow x \in {\frac{2 π}{3}, \frac{4 π}{3}}$ $cos(x) = -1/2 <=> x in {(2pi)/3, (4pi)/3}$
$cos (x) = \frac{1}{2} \Leftrightarrow x \in {\frac{π}{3}, \frac{5 π}{3}}$ $cos(x) = 1/2 <=> x in {pi/3, (5pi)/3}$

So, putting those together, we get our final result:

$x \in {\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}}$ $x in {pi/3, (2pi)/3, (4pi)/3, (5pi)/3}$

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How do you solve $2 = 4 {cos}^{2} x + 1$ $2=4cos^2x+1$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve $2 = 4 {cos}^{2} x + 1$ $2=4cos^2x+1$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?