How do you find all solutions of the equation $cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) = 0$ $cos(x+(3pi)/4)-cos(x-(3pi)/4)=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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How do you find all solutions of the equation $cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) = 0$ $cos(x+(3pi)/4)-cos(x-(3pi)/4)=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2017-01-09T15:23:59

$0, π, 2 π$ $0, pi, 2pi$

Explanation:

Use trig identity:
$cos a - cos b = - 2 sin (\frac{a + b}{2}) sin (\frac{a - b}{2})$ $cos a - cos b = - 2sin ((a + b)/2)sin ((a - b)/2)$
In this case:
(a + b) = 2x --> $sin (\frac{a + b}{2}) = sin x$ $sin ((a + b)/2) = sin x$
$(a - b) = \frac{6 π}{4}$ $(a - b) = (6pi)/4$ --> $sin (\frac{a - b}{2}) = sin (\frac{3 π}{4})$ $sin ((a - b)/2) = sin ((3pi)/4)$
The equation f(x) becomes:
$f (x) = cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) =$ $f(x) = cos (x + (3pi)/4) - cos (x - (3pi)/4) =$
$= - 2 sin x . sin (\frac{3 π}{4}) = 0$ $= - 2sin x.sin ((3pi)/4) = 0$
Since $sin (\frac{3 π}{4}) = \frac{\sqrt{2}}{2}$ $sin ((3pi)/4) = sqrt2/2$ , then:
$f (x) = - \sqrt{2} . sin x = 0$ $f(x) = - sqrt2.sin x = 0$
Unit circle -->
sin x = 0 --> $x = 0, x = π, x = 2 π$ $x = 0, x= pi, x = 2pi$
Answers for $(0, 2 π)$ $(0, 2pi)$
$0, π, 2 π$ $0, pi, 2pi$

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How do you find all solutions of the equation $cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) = 0$ $cos(x+(3pi)/4)-cos(x-(3pi)/4)=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Explanation:

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How do you find all solutions of the equation $cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) = 0$ $cos(x+(3pi)/4)-cos(x-(3pi)/4)=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?