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Factor the trigonometric expression: cos^2(x)sec(x)-4cos(x)sec(x)-21sec(x)
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Wouldn't sec(x) be a common factor?
that's what i thought... but for the input it has an empty box* empty box *empty box
and I'm not sure what the 3 would be..
If you factor out a sec(x), you get: sec(x) [cos^2 (x) - 4 cos(x) - 21} But that second expression may be factorable again.
oh ok! thank you! could you factor the second one into 4cos(x) [ cos(x)]? but then where would the 21 go
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cosx=7 cosx =-3 right? and 0
sub u for cos then solve like a regular equation?
secx (cos x - 7)(cos x +3)
Perfect!
oh my gosh, thank you!!!!
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welcome.
Thank you to both of you @Easyaspi314 and @sarsteele !!!
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