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Trigonometrical equation 2

Solve the following for 0° ≤ x ≤ 360°:
5secx +3cosecx = 0
I've tried substituting things but I can't seem to get the two parts into the same form, which I'm assuming you have to do? Any help?
Original post by ZoeM0410
Solve the following for 0° ≤ x ≤ 360°:
5secx +3cosecx = 0
I've tried substituting things but I can't seem to get the two parts into the same form, which I'm assuming you have to do? Any help?

you just have to rearange the equation and use
secx=1/cosx cosecx=1/sinx and tanx=(sinx)/(cosx)
Original post by ZoeM0410
Solve the following for 0° ≤ x ≤ 360°:
5secx +3cosecx = 0
I've tried substituting things but I can't seem to get the two parts into the same form, which I'm assuming you have to do? Any help?


Have you considered the following?

sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}

csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}
Original post by ZoeM0410
Solve the following for 0° ≤ x ≤ 360°:
5secx +3cosecx = 0
I've tried substituting things but I can't seem to get the two parts into the same form, which I'm assuming you have to do? Any help?

So
5/cosx = - 3/sinx
I'd Multiply through by sinx.
5sinx / cosx = -3
Notice that sinx/cosx = tanx

Are you ok from here?
Reply 4
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ZoeM0410
OP
Thanks ! That really helps, I've got it now ! :biggrin:

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