Question

Hey guys, this is a seemingly easy trig question but I cant seem to get the right answer to sumbit online. The question is (Cosx)^2-1.6cosx-0.57=0. Solve on the interval [0,2pi]. So I did the quadratic and I get x=1.9 and x=-0.3. I have tryed a bunch of steps from here but none are getting the answer. Any insight would be greatly appreciated, thanks!

Answer 1

you have 2 solutions right? cos x = 1.9 and cos x = -0.3
Pay attention, your quadratic is cos^2 ..... not x ^2. therefore, the solutions must be respect to cos x
reject the first value, because cos of an angle cannot exceed 1
So, you just have cos x = -0.3 which leads to x = 1.875

Answer 2

let me try, I forgot which mode I set up on my Ti 83, so, not sure that value is in radian or degree

Answer 3

http://www.felderbooks.com/papers/ti.html

Answer 4

ok, it says x = 107.46 degree.

Answer 5

thank you mathematician jdoe0001 :)

Answer 6

Thanks a lot for the help, and yup Ive tried the 107.46, it is not the right answer.

Answer 7

ok

Answer 8

I am useless here. hehehe... @jdoe0001

Answer 9

hmm, lemme do a quickie

Answer 10

\(\bf cos^2(x)-1.6cos(x)-0.57=0\\
\textit{let's multiply both sides by 10}\\
10cos^2(x)-16cos(x)-57=0\\ \quad \\
x = \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\\ \quad \\\implies cos(x) = \cfrac{ - (-16) \pm \sqrt { (-16)^2 -4(10)(-57)}}{2(10)}\\ \quad \\
cos(x) = \cfrac{ 16 \pm \sqrt { 256 +2280}}{20}\implies
cos(x) = \cfrac{ 16 \pm \sqrt { 2536}}{20}\\ \quad \\

cos(x) \implies
\begin{cases}
\cfrac{16+50.35871325}{20} \approx 3.32\\ \quad \\
\bf \cfrac{16-50.35871325}{20} \approx 1.72
\end{cases}\)

Answer 11

well.. negative on the last one... \(\bf cos(x) \implies
\begin{cases}
\cfrac{16+50.35871325}{20} \approx 3.32\\ \quad \\
\bf \cfrac{16-50.35871325}{20} \approx -1.72
\end{cases}\)