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How do you solve $(4 {cos}^{2} θ - 2) + (6 {cos}^{2} θ - 3) = 1$ $(4cos^2theta - 2) + (6cos^2theta - 3) = 1$ ?

1 Answer

Aug 5, 2015

Within the range $θ ε [0, π]$ $theta epsilon [0,pi]$
$XXXX$ $color(white)("XXXX")$ $θ = 0.684719$ $theta = 0.684719$ radians
or
$XXXX$ $color(white)("XXXX")$ $θ = 2.456873$ $theta = 2.456873$ radians

Explanation:

$(4 {cos}^{2} (θ) - 2) + (6 {cos}^{2} (θ) - 3) = 1$ $(4cos^2(theta) - 2)+ (6cos^2(theta)-3) = 1$

Is equivalent to
$XXXX$ $color(white)("XXXX")$ $10 {cos}^{2} (θ) = 6$ $10cos^2(theta) = 6$

$\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ ${cos}^{2} (θ) = 0.6$ $cos^2(theta) = 0.6$

$\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ $cos (θ) = \pm \sqrt{0.6}$ $cos(theta) =+- sqrt(0.6)$

$\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ $θ = arccos (\sqrt{0.6}) = 0.684719$ $theta = arccos(sqrt(0.6)) = 0.684719$
or
$\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ $θ = arccos (- \sqrt{0.6}) = 2.45873$ $theta = arccos(-sqrt(0.6)) = 2.45873$

(Sorry, but I couldn't find anything prettier.)

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