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Trig equations

When I get solutions like cot(x)=2 I just use 1/tan(x)=2/1 and flip them both, but I don't know what to do when it's something like cosec(x)=0, should I just realise that cosec(x) can never =0 because of the graph? And in that case when I get cot(x)=0, should I just do cos(x)/sin(x)=0?
Reply 1
Avatar for nerak99
nerak99
13
If you got cot(x)=0 then what you suggest does work. Hence x=90. This is quite interesting actually. Never thought of doing that before.
Reply 2
Avatar for Zacken
Zacken
22
Original post by white_o
When I get solutions like cot(x)=2 I just use 1/tan(x)=2/1 and flip them both, but I don't know what to do when it's something like cosec(x)=0, should I just realise that cosec(x) can never =0 because of the graph? And in that case when I get cot(x)=0, should I just do cos(x)/sin(x)=0?


csc⁡x=0\csc x = 0 has no solutions, like you said.

cot⁡x=0\cot x = 0 occurs precisely when cos⁡xsin⁡x=0⇒cos⁡x=0\frac{\cos x}{\sin x} = 0 \Rightarrow \cos x = 0.
(edited 7 years ago)
Reply 3
Avatar for white_o
white_o
OP
17
Original post by nerak99
If you got cot(x)=0 then what you suggest does work. Hence x=90. This is quite interesting actually. Never thought of doing that before.

Okay, thanks!

Original post by Zacken
csc⁡x=0\csc x = 0 has no solutions, like you said.

cot⁡x=0\cot x = 0 occurs precisely when cos⁡xsin⁡x=0⇒cos⁡x=0\frac{\cos x}{\sin x} = 0 \Rightarrow \cos x = 0.

Right, thank you :smile:
Reply 4
Avatar for Zacken
Zacken
22
Original post by white_o

Right, thank you :smile:


No problem. :smile:

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