How do you solve $sin(x+pi/4)+sin(x-pi/4)=1$ over the interval $(0,2pi)$ ?

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How do you solve $sin(x+pi/4)+sin(x-pi/4)=1$ over the interval $(0,2pi)$ ?

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Answer 1 · 2015-04-17T03:50:53

Use the trig identity: $color(blue)(sin (a + b) + sin (a - b) = 2sin a*cos b)$

$f(x) = 2*sin x*cos (pi/4) - 1 = 0$ ,

since $color(blue)(cos (pi/4) = (sqrt2)/2$

$f(x) = (sqrt2*sin x) - 1 = 0$

$sin x = 1/sqrt2 = (sqrt2)/2 -->$

$color(red)(x = pi/4 and 3pi/4$ (inside interval $0 - 2pi$ )

Check:
$x = pi/4 --> x + pi/4 = pi/2 --> sin (x + pi/4) = 1; cos (x + pi/4) = 0 --> f(x) = 1 - 1 = 0$ . Correct.
$x = 3pi/4 --> (x + pi/4) = pi --> sin pi = 0; cos pi = -1 --> f(x) = 1 - 1 = 0.$ Correct

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How do you solve $sin(x+pi/4)+sin(x-pi/4)=1$ over the interval $(0,2pi)$ ?

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