How do you solve for x in $2 cos (3 x - 1) = 0$ $2cos(3x-1)=0$ ?

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How do you solve for x in $2 cos (3 x - 1) = 0$ $2cos(3x-1)=0$ ?

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Answer 1 · 2015-02-12T11:53:00

If $2 cos (3 x - 1) = 0$ $2cos(3x-1)=0$ then so is $cos (3 x - 1)$ $cos(3x-1)$

If a $cos θ = 0$ $cos theta=0$ then $θ = 90^{0}$ $theta=90^0$ plus any multiple of $180^{0}$ $180^0$
(the cosine function crosses the $x$ $x$ -axis twice every period)

So we have
$3 x - 1 = 90^{0} + k \cdot 180^{0}$ $3x-1=90^0+k*180^0$

$\to 3 x = 91^{0} + k \cdot 180^{0}$ $->3x=91^0+k*180^0$

$\to x = \frac{91^{0}}{3} + k \cdot \frac{180^{0}}{3}$ $->x=91^0/3+k*180^0/3$

Answer:
$x = {(30 \frac{1}{3})}^{0} + k \cdot 60^{0}$ $x=(30 1/3) ^0 +k*60^0$

Extra:
If you work in radians (you didn't say), then solving would go along the same lines, only then you would work with

$3 x - 1 = \frac{π}{2} + k . π$ $3x-1=pi/2 +k.pi$

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How do you solve for x in $2 cos (3 x - 1) = 0$ $2cos(3x-1)=0$ ?

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