SOLUTION: Find all solutions of the equation in the interval [0,2pi) sin (x+7pi) - sinx - sqrt.2 =0

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Question 674447: Find all solutions of the equation in the interval [0,2pi)
sin (x+7pi) - sinx - sqrt.2 =0

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find all solutions of the equation in the interval [0,2pi)
sin (x+7pi) - sinx - sqrt.2 =0
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sin addition formula: sin(a+b)=sin a cos b+cos a sin b
sin(x+7π)=sin x cos 7π+cos x sin 7π)
cos 7π=-1
sin 7π=0
sin x cos 7π+cos x sin 7π)
sin x*(-1)+cos x*(0)
sin(x+7π)=-sin x
..
sin (x+7π) - sinx - sqrt.2 =0
-sin x-sin x-√2=0
-2sin x=√2
sin x=-√2/2
x=5π/4 and 7π/4 (in quadrants III and IV where sin<0)