How do you solve $sin (2 x) = - cos (2 x)$ $sin(2x)=-cos(2x)$ ?

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How do you solve $sin (2 x) = - cos (2 x)$ $sin(2x)=-cos(2x)$ ?

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Answer 1 · 2017-06-30T18:05:45

$x = \frac{3 π}{8} + k π$ $x = (3pi)/8 + kpi$
$x = \frac{7 π}{8} + k π$ $x = (7pi)/8 + kpi$

Explanation:

sin 2x + cos 2x = 0
Use trig identity: $sin a + cos a = \sqrt{2} cos (a - \frac{π}{4})$ $sin a + cos a = sqrt2cos (a - pi/4)$
In this case:
$sin 2 x + cos 2 x = \sqrt{2} cos (2 x - \frac{π}{4}) = 0$ $sin 2x + cos 2x = sqrt2cos (2x - pi/4) = 0$
$cos (2 x - \frac{π}{4}) = 0$ $cos (2x - pi/4) = 0$
Unit circle gives 2 solutions:
a. $2 x - \frac{π}{4}) = \frac{π}{2} + 2 k π$ $2x - pi/4) = pi/2 + 2kpi$ --> $2 x = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4} + 2 k π$ $2x = pi/2 + pi/4 = (3pi)/4 + 2kpi$ -->
$x = \frac{3 π}{8} + k π$ $x = (3pi)/8 + kpi$ .
b. $(2 x - \frac{π}{4}) = \frac{3 π}{2} + 2 k π$ $(2x - pi/4) = (3pi)/2 + 2kpi$ --> $2 x = \frac{3 π}{2} + \frac{π}{4} = \frac{7 π}{4} + 2 k π$ $2x = (3pi)/2 + pi/4 = (7pi)/4 + 2kpi$ --> $x = \frac{7 π}{8} + k π$ $x = (7pi)/8 + kpi$

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How do you solve $sin (2 x) = - cos (2 x)$ $sin(2x)=-cos(2x)$ ?

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