# Trig equation

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#1
solve tan^2x=secxtanx for 0°≤x≤360°
I got sin^2x-sinx=0 which gives x=0 or 1
which I know is correct but then the solutions are x=0°,180°,360°
I don't understand why x=90° isn't a possible solution?
0
1 month ago
#2
(Original post by learningizk00l)
solve tan^2x=secxtanx for 0°≤x≤360°
I got sin^2x-sinx=0 which gives x=0 or 1
which I know is correct but then the solutions are x=0°,180°,360°
I don't understand why x=90° isn't a possible solution?
What do you get subbing it into the original equation?
1
#3
(Original post by mqb2766)
What do you get subbing it into the original equation?
oh yeah tan undefined at x=90° and also cos(90)=0 so would have 1/0 again undefined
Thanks!
0
1 month ago
#4
(Original post by learningizk00l)
oh yeah tan undefined at x=90° and also cos(90)=0 so would have 1/0 again undefined
Thanks!
Yes. Im presuming you multiplied through by cos() squared in your solution. You should always be aware that this could be zero and state that the derivation is only true if its non-zero.
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