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Question #a6825

1 Answer

Feb 17, 2018

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Explanation:

Instead of $cos (2 x) = 0$ $cos(2x)=0$ , think of the equation as $cos (θ) = 0$ $cos(theta)=0$ , where $θ = 2 x$ $theta=2x$

Since $cos θ = 0$ $costheta=0$ at $θ = \frac{π}{2}$ $theta=pi/2$ and $θ = \frac{3 π}{2}$ $theta=(3pi)/2$ , then by substitution,

$cos (2 x) = 0$ $cos(2x)=0$ at $2 x = \frac{π}{2}$ $2x=pi/2$ and $2 x = \frac{3 π}{2}$ $2x=(3pi)/2$

Solve each equation separately

$2 x = \frac{π}{2}$ $2x=pi/2$
$x = \frac{π}{2} \times \frac{1}{2}$ $x=pi/2xx1/2$
$x = \frac{π}{4}$ $x=pi/4$

$2 x = \frac{3 π}{2}$ $2x=(3pi)/2$
$x = \frac{3 π}{2} \times \frac{1}{2}$ $x=(3pi)/2xx1/2$
$x = \frac{3 π}{4}$ $x=(3pi)/4$

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