Question

sec^4theta-8sec^2(theta)*tan(theta)+16tan^2(theta)=0 if theta=?

Answer 1

\[\sec^2(\theta)-8\sec^2(\theta)\tan(\theta)+16\tan^2(\theta)=0\]
let's just write in terms of \[\tan(\theta)\]
first of all recall that
\[\sin^2(\theta)+\cos^2(\theta)=1\]
dividing both sides by \[\cos^2(\theta)\] will give us the identity we need
\[\tan^2(\theta)+1=\sec^2(\theta)\]
so now we can write our equation as
\[(\tan^2(\theta)+1)^2-8(\tan^2(\theta)+1)\tan(\theta)+16\tan^2(\theta)=0\]
now lets multiply a little
\[\tan^4(\theta)+2\tan^2(\theta)+1-8\tan^3(\theta)-8\tan(\theta)+16\tan^2(\theta)=0\]
if any combine like terms
\[\tan^4(\theta)-8\tan^3(\theta)+18\tan^2(\theta)-8\tan(\theta)+1=0\]
now lets use a substitution to make this easier on our eyes
let \[u=\tan(\theta)\]
so we have
\[u^4-8u^3+18u^2-8u+1=0\]
now we need to find the zeros of this polynomial
\[1-8+18-8+1=-7+18-7=-14+18=4 \neq 0\]
\[1+8+18+8+1 \neq 0\]
so we have that 1 or -1 don't work
so that means we have irrational solutions for this polynomial

Answer 2

its ans is \[\Theta=5\pi/12 \Theta=\Pi /12, \]

Answer 3

let t=tan(theta), the given equation became t^4-8t^3+18t^2+1=0
x=t+1/t
=)x^2-8x+16=0
x=4 t=2+-\[\sqrt{3}\]

Answer 4

its very simple problem to solv

Answer 5

wait our polynomials are different
i have an extra term
let me look at this

Answer 6

\[u^4-8u^3+18u^2-8u+1=0 \]
i see this
let me know if i made a mistake
i know in my first line i typed it wrong but i fixed it when i was working on it above