How do I solve 1=2cos(3X-5pi) for (0,2pi)?

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How do I solve 1=2cos(3X-5pi) for (0,2pi)?

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

2 Answers

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Answer 1 · 2018-02-15T20:25:06

The values of x being $\frac{16 π}{9} and \frac{20 π}{9}$ $(16pi)/9 and (20pi)/9$ form the valid solutions.

Explanation:

Given:
$1 = 2 cos (3 x - 5 π)$ $1=2cos(3x-5pi)$
for $(0, 2 π)$ $(0,2pi)$
Interchanging lhs and rhs
$2 cos (3 x - 5 π) = 1$ $2cos(3x-5pi)=1$

Dividing by 2
$cos (3 x - 5 π) = \frac{1}{2}$ $cos(3x-5pi)=1/2$
We know that $cos (\frac{π}{6}) = \frac{1}{2}$ $cos(pi/6)=1/2$ is the fundamental solution
Comparing
$cos (3 x - 5 π) = \frac{1}{2}$ $cos(3x-5pi)=1/2$ and $cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3)=1/2$
we have, $3 x - 5 π = \frac{π}{3}, 2 π - \frac{π}{3}$ $3x-5pi=pi/3,2pi-pi/3$ for the range $(0, 2 π)$ $(0,2pi)$

$3 x - 5 π = \frac{π}{3}$ $3x-5pi=pi/3$
Adding $5 π$ $5pi$
$3 x = 5 π + \frac{π}{3} = \frac{31 π}{6}$ $3x=5pi+pi/3=(31pi)/6$
Dividing by 3
$x = (\frac{1}{3}) \frac{16 π}{3} = \frac{16 π}{9}$ $x=(1/3)(16pi)/3=(16pi)/9$

$3 x - 5 π = 2 π - \frac{π}{3} = \frac{5 π}{3}$ $3x-5pi=2pi-pi/3=(5pi)/3$
Adding $5 π$ $5pi$
$3 x = 5 π + \frac{5 π}{3} = \frac{20 π}{3}$ $3x=5pi+(5pi)/3=(20pi)/3$
Dividing by 3
$x = \frac{20 π}{9}$ $x=(20pi)/9$

The values of x being $\frac{16 π}{9} and \frac{20 π}{9}$ $(16pi)/9 and (20pi)/9$ form the valid solutions.

Answer 2 · 2018-02-16T23:48:11

$\frac{2 π}{9}; \frac{4 π}{9}$ $(2pi)/9; (4pi)/9$

Explanation:

$2 cos (3 x - 5 π) = 2 cos (3 x - π) = 1$ $2cos (3x - 5pi) = 2cos (3x - pi) = 1$
$cos (3 x - π) = \frac{1}{2}$ $cos (3x - pi) = 1/2$
Trig table and unit circle give 2 solutions:
$3 x - π = \pm \frac{π}{3}$ $3x - pi = +- pi/3$
a. $3 x - π = \frac{π}{3}$ $3x - pi = pi/3$ --> $3 x = π + \frac{π}{3} = \frac{4 π}{3}$ $3x = pi + pi/3 = (4pi)/3$
$x = \frac{4 π}{9}$ $x = (4pi)/9$
b. $3 x - π = - \frac{π}{3}$ $3x - pi = - pi/3$ --> $3 x = π - \frac{π}{3} = \frac{2 π}{3}$ $3x = pi - pi/3 = (2pi)/3$
$x = \frac{2 π}{9}$ $x = (2pi)/9$

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How do I solve 1=2cos(3X-5pi) for (0,2pi)?

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

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How do I solve 1=2cos(3X-5pi) for (0,2pi)?

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

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Explanation:

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