How do you solve $sin 2 x = cos 2 x$ $sin2 x = cos2 x$ over the interval 0 to 2pi?

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How do you solve $sin 2 x = cos 2 x$ $sin2 x = cos2 x$ over the interval 0 to 2pi?

1 Answer

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Answer 1 · 2016-03-16T09:20:37

$\frac{π}{8} and 5 \frac{π}{8}$ $pi/8 and 5pi/8$ .

Explanation:

$sin (π + 2 x) = - sin 2 x and cos (π + 2 x) = - cos 2 x$ $sin(pi+2x) =-sin 2x and cos(pi+2x) = -cos 2x$ .

The solutions are 2x = $\frac{π}{4} and 2 x = π + \frac{π}{4} = 5 \frac{π}{4}$ $pi/4 and 2x = pi+pi/4 = 5pi/4$ .
$x = \frac{π}{8} and 5 \frac{π}{8}$ $x = pi/8 and 5pi/8$ ....

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How do you solve $sin 2 x = cos 2 x$ $sin2 x = cos2 x$ over the interval 0 to 2pi?

1 Answer

Explanation:

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How do you solve sin2 x = cos2 xsin2x=cos2xsin2 x = cos2 x over the interval 0 to 2pi?

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Explanation:

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How do you solve $sin 2 x = cos 2 x$ $sin2 x = cos2 x$ over the interval 0 to 2pi?