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Mathematics 24 Online
OpenStudy (anonymous):

HELP PLEASE!!!!! solve 6cos theta+2=0

OpenStudy (anonymous):

\[6\cos \theta +2=\]

OpenStudy (anonymous):

a=109.5 b=30.7 c=28.6 d=70.5

OpenStudy (anonymous):

what's theta?

OpenStudy (anonymous):

θ= theta

OpenStudy (anonymous):

There's gotta be another piece of information in that problem...can you look again?

OpenStudy (anonymous):

6 cos theta + 2 = 0

OpenStudy (anonymous):

\[6\cos(\theta) = -2\] \[\cos(\theta) = \frac{-1}{3} \implies \theta = \cos^{-1}(\frac{-1}{3})\] \[\theta = \cos^{-1}(\frac{1}{3}) \implies 70.73 \approx 70.5\]

OpenStudy (anonymous):

d=70.5

OpenStudy (anonymous):

thank you could you also help me with a few others??

OpenStudy (anonymous):

Not sure but I will try..

OpenStudy (anonymous):

\[\sin ^{2}x+3 sin x+2=0\]

OpenStudy (anonymous):

u there @waterineyes

OpenStudy (anonymous):

Here its 2:11 am.. I am falling asleep.. Here you can factorize it: \[\sin ^{2}x+3 \sin x+2=0 \implies (sinx+2)(sinx + 1) = 0\] Put them separately equal to 0 and find sinx from there..

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