Solving Trig Equation secx + cosx = tanx sinx - 1
- Thread starter mathcholo
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D
Deleted member 4993
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secx+cosx=tanxsinx-1
Please help I can't figure it out
What do you need to do?
Prove identity?
Solve for 'x'?
What are your thoughts?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions.
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Sorry, I'm new here. We are supposed to set the equation equal to zero to see what X equals within (0, 2pi)
I tried it two ways:
tan^2x-5tanx+5=0
(tanx-3)(tanx-2)=0
(tanx-3)=0 (tanx-2)=0
tanx=3 tanx=2
but then what? it doesn't seem to work out because in class the answer seemed to come out to be like tanx=sqrt. 3
then it the final answer would be pi/3 and 4pi/3.
My other method:
tan^2x-5tanx+5=0
tan^2x-5tanx=-5
tanx(tanx-5)=-5
but then I'm confused how i can set tanx=-5 or (tanx-5)=-5
Sorry for not reading the terms, will read them now!
I tried it two ways:
tan^2x-5tanx+5=0
(tanx-3)(tanx-2)=0
(tanx-3)=0 (tanx-2)=0
tanx=3 tanx=2
but then what? it doesn't seem to work out because in class the answer seemed to come out to be like tanx=sqrt. 3
then it the final answer would be pi/3 and 4pi/3.
My other method:
tan^2x-5tanx+5=0
tan^2x-5tanx=-5
tanx(tanx-5)=-5
but then I'm confused how i can set tanx=-5 or (tanx-5)=-5
Sorry for not reading the terms, will read them now!
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- Feb 4, 2004
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How does this relate to the original equation...?
D
Deleted member 4993
Guest
But you started with:
secx + cosx = tanxsinx - 1
multiplying both sides by cos(x) [x <> π/2]
1 + cos2(x) = sin2(x) - cos(x)
and continue....