(tanΘ - 1) (secΘ - 1) = 0?
Solve equation on the interval 0 ≤ Θ ≤ 2pi.
I think I have to use identities for this, but I can't figure it what I have to do with the negative 1's on the inside of the parantheses. My textbook only has like two examples and this case isn't shown in any of them.
So far, I have this written down:
(sinΘ -1)/(cosΘ - 1) * (Θ - 1)/(cosΘ - 1)
That's because tanΘ = sinΘ/cos Θ and secΘ = 1/cosΘ
Is that right? If not, what are the proper steps? If it IS right, can you please show me the following steps? I'm sooo confused!
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dude, ur killing urself. u have two factors. if one of them is 0 the whole thing is 0.
tan - 1 = 0
tan = 1
x = pi/4 (Quad 1) and 5pi/4 (Quad 3) (where tangent is +)
sec - 1 = 0
sec = 1
sec = 1/cos
1/1 = 1
cos =1
x = 0 and (2pi if its in your interval, but usually books leave that out since 2pi is coterminal with 0)
good luck
(tan o + a million) * (sec 0 - .5) = 0 answer from 0 to 360 Tan theta +a million = 0 and Sec theta -.5 = 0 Tan theta = -a million => a hundred thirty five, 315 : Tan function has a era of one hundred eighty stages. sec theta = .5 => cos theta = 2 => no longer plausible root
tan(x)=1 or sec(x)=1
tan(x)=1 or cos(x)=1
tan(x)=1 ---> (pi/4) , pi +(pi/4)
cos(x)=1 ---> x=0, 2pi
Therefore, all the solns in the interval are
x= 0, pi/4 , 5pi/4 , 2pi