How do i solve 1=2cos(3x-5pi) for [0,2)?

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How do i solve 1=2cos(3x-5pi) for [0,2)?

$1 = 2 cos (3 x - 5 π)$ $1=2cos(3x-5pi)$ for $[0, 2 π) ?$ $[0,2pi)?$

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Answer 1 · 2018-02-14T02:41:46

$x = \pm \frac{2 π}{9}$ $x = +- (2pi)/9$ , or
$x = \pm 40^{\circ}$ $x = +- 40^@$

Explanation:

$cos (3 x - 5 π) = \frac{1}{2}$ $cos (3x - 5pi) = 1/2$
$cos (3 x - 5 π) = cos (3 x - π) = - cos 3 x$ $cos (3x - 5pi) = cos (3x - pi) = - cos 3x$ -->
$cos 3 x = - \frac{1}{2}$ $cos 3x = - 1/2$
Trig table and unit circle give 2 solutions:
$3 x = \pm \frac{2 π}{3}$ $3x = +- (2pi)/3$
$x = \pm \frac{2 π}{9}$ $x = +- (2pi)/9$
Check by calculator:
$x = \frac{2 π}{9} = 40^{\circ}$ $x = (2pi)/9 = 40^@$ --> $2 cos (3 x - π) = 2 cos (120 - 180) =$ $2cos (3x - pi) = 2cos (120 - 180) =$
$= 2 cos (- 60) = 2 cos 60 = 2 (\frac{1}{2}) = 1$ $= 2cos (-60) = 2cos 60 = 2(1/2) = 1$ Proved.

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How do i solve 1=2cos(3x-5pi) for [0,2)?

$1 = 2 cos (3 x - 5 π)$ $1=2cos(3x-5pi)$ for $[0, 2 π) ?$ $[0,2pi)?$

1 Answer

Explanation:

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How do i solve 1=2cos(3x-5pi) for [0,2)?

1=2cos(3x-5pi)1=2cos(3x−5π)1=2cos(3x-5pi) for [0,2pi)? [0,2π)?[0,2pi)?

1 Answer

Explanation:

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$1 = 2 cos (3 x - 5 π)$ $1=2cos(3x-5pi)$ for $[0, 2 π) ?$ $[0,2pi)?$