SOLUTION: Solve the multi-angle below sin(2x)= - square root of 2 / 2

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Question 873027: Solve the multi-angle below
sin(2x)= - square root of 2 / 2
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If sin%282x%29=+-sqrt%282%29%2F2 It could be that 2x=-pi%2F4 or 2x=-3pi%2F4 (for angles AOP and AOQ ).
2x could also be any other angle differing from those by a multiple of 2pi .
Starting from 2x=-pi%2F2+%2B-+pi%2F4=system%282x=-pi%2F4%2C%22or%22%2C2x=-3pi%2F4%29 ,
or from 2x=3pi%2F2+%2B-+pi%2F4=system%282x=5pi%2F4%2C%22or%22%2C2x=7pi%2F4%29
a way of expressing all those angle with one formula is
2x=%282k%2Api-pi%2F2%29+%2B-+pi%2F4=8k%2Api%2F4-2pi%2F4+%2B-+pi%2F4=%288k-2+%2B-+1%29pi%2F4 for any integer k .
So 2x=%288k-2+%2B-+1%29pi%2F4 ---> highlight%28x=%288k-2+%2B-+1%29pi%2F8%29 for any integer k .
That gives you infinite pairs of co-terminal angles:
For k=-1 we have x=-11pi%2F8 and x=-9pi%2F8 .
For k=0 we have x=-3pi%2F8 and x=-pi%2F8 .
For k=1 we have x=5pi%2F8 and x=7pi%2F8 .
For k=2 we have x=13pi%2F8 and x=15pi%2F8 .
There are infinitely more pairs.