anonymous
  • anonymous
sin(y/2 + pi/3) = 1/root2 solve for 0
Mathematics
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anonymous
  • anonymous
sin(y/2 + pi/3) = 1/root2 solve for 0
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
myininaya got this one i bet
myininaya
  • myininaya
\[\sin(u)=\frac{\sqrt{2}}{2}\] when \[u=\frac{\pi}{4}+2 n \pi\] \[u=\frac{3 \pi}{4 }+2n \pi\] n is an integer
myininaya
  • myininaya
so that means we have \[\frac{y}{2}+\frac{\pi}{3}=\frac{\pi}{4}+2 n \pi\] \[\frac{y}{2}+ \frac{\pi}{3}=\frac{3 \pi}{4}+2 n \pi\]

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myininaya
  • myininaya
so solve both of these for y
myininaya
  • myininaya
\[y+\frac{2 \pi}{3}=\frac{\pi}{2}+4 n \pi\] \[y+\frac{2 \pi}{3}=\frac{3 \pi}{2}+4 n \pi\]
myininaya
  • myininaya
\[y=\frac{\pi}{2}-\frac{2 \pi}{3} + 4 n \pi\] \[y=\frac{3 \pi}{2}-\frac{2 \pi}{3}+4 n \pi\]
myininaya
  • myininaya
\[y=\frac{3 \pi - 4 \pi}{6}+4 n \pi\] \[y=\frac{9 \pi -4 \pi}{6}+4 n \pi\]
myininaya
  • myininaya
\[y=\frac{-\pi}{6}+4 n \pi\] \[y=\frac{5 \pi}{6}+4 n \pi\]
myininaya
  • myininaya
for first equation: \[n=0 => y=\frac{-\pi}{6} <0\] \[n=1 => y=-\frac{\pi}{6}+4 \pi=\frac{-\pi+24 \pi}{6}=\frac{23 \pi}{6}<4 \pi\] \[n=2 => y=\frac{-\pi}{6}+8 \pi=\frac{47 \pi}{6} >4 \pi\] for second equation: \[n=0 => y=\frac{5 \pi}{6}<4 \pi\] \[n=-1=> \frac{5 \pi}{6}-4\pi=\frac{5 \pi-24 \pi}{6}=\frac{-19 \pi}{6}<0\]
myininaya
  • myininaya
so our solutions are... \[y=\frac{ 23 \pi}{6}, \frac{5 \pi}{6}\]
myininaya
  • myininaya
for the interval of y given

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