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Mathematics 8 Online
OpenStudy (anonymous):

Find all solutions to the equation. (sin x)(cos x) = 0

OpenStudy (anonymous):

anyone help me?!?!:(

jimthompson5910 (jim_thompson5910):

Hint: if (sin x)(cos x) = 0 then sin(x) = 0 or cos(x) = 0

jimthompson5910 (jim_thompson5910):

now use the unit circle to figure out when sin(x) = 0 or when cos(x) = 0

OpenStudy (anonymous):

@jim_thompson5910 so {(pi/2)+npi|n=0,1,2.....}? |

jimthompson5910 (jim_thompson5910):

well if n = 0, then pi/2+npi turns into pi/2 but x = 0 is a solution to sin(x) = 0

jimthompson5910 (jim_thompson5910):

so it's close, but not quite there

jimthompson5910 (jim_thompson5910):

does that make sense?

OpenStudy (anonymous):

@jim_thompson5910 yeaaa so what would be the correct answer?

jimthompson5910 (jim_thompson5910):

a better answer would be {n*pi/2 | n = 0, 1, 2, ...} because notice now that when n = 0, x is x = 0 when n = 1, x = pi/2 when n = 2, x = pi etc etc

jimthompson5910 (jim_thompson5910):

pretty much, any nonnegative multiple of pi/2 is a solution

OpenStudy (anonymous):

@jim_thompson5910 {(pi/2)+npi,npi|n=0,1,2 would this be right then? {(pi/2)+npi|n=0,1,2} or this?

jimthompson5910 (jim_thompson5910):

oh i see what they did, they broke it up

jimthompson5910 (jim_thompson5910):

so yes it would be {(pi/2)+npi,npi|n=0,1,2, ... }

jimthompson5910 (jim_thompson5910):

one is the solution set to cos(x) = 0, the other is the solution set to sin(x) = 0

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