Question

sin(y/2 + pi/3) = 1/root2
solve for 0 14 years ago

Answer 1

myininaya got this one i bet

Answer 2

\[\sin(u)=\frac{\sqrt{2}}{2}\]
when
\[u=\frac{\pi}{4}+2 n \pi\]
\[u=\frac{3 \pi}{4 }+2n \pi\]
n is an integer

Answer 3

so that means we have
\[\frac{y}{2}+\frac{\pi}{3}=\frac{\pi}{4}+2 n \pi\]
\[\frac{y}{2}+ \frac{\pi}{3}=\frac{3 \pi}{4}+2 n \pi\]

Answer 4

so solve both of these for y

Answer 5

\[y+\frac{2 \pi}{3}=\frac{\pi}{2}+4 n \pi\]
\[y+\frac{2 \pi}{3}=\frac{3 \pi}{2}+4 n \pi\]

Answer 6

\[y=\frac{\pi}{2}-\frac{2 \pi}{3} + 4 n \pi\]
\[y=\frac{3 \pi}{2}-\frac{2 \pi}{3}+4 n \pi\]

Answer 7

\[y=\frac{3 \pi - 4 \pi}{6}+4 n \pi\]
\[y=\frac{9 \pi -4 \pi}{6}+4 n \pi\]

Answer 8

\[y=\frac{-\pi}{6}+4 n \pi\]
\[y=\frac{5 \pi}{6}+4 n \pi\]

Answer 9

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:
\[n=0 => y=\frac{5 \pi}{6}<4 \pi\]
\[n=-1=> \frac{5 \pi}{6}-4\pi=\frac{5 \pi-24 \pi}{6}=\frac{-19 \pi}{6}<0\]

Answer 10

so our solutions are...
\[y=\frac{ 23 \pi}{6}, \frac{5 \pi}{6}\]

Answer 11

for the interval of y given