4. Find all values of x in [0, 2π] that satisfy the equation: 5 cos x − 2 sin^2 x = 1

Dec 2019
8
0
Windsor Ontario Canada
4. Find all values of x in [0, 2π] that satisfy the equation: 5 cos x − 2 sin^2 x = 1
Answer: x = π / 3 , 5π / 3

I'm not entirely sure what steps need to be taken to solve this problem. I've seen simplier problems withlike such 2cosx - 1 = 0
where you just solve for cos x and relate that to the unit circle
 
Dec 2019
8
0
Windsor Ontario Canada
Do you mind writing all the steps for the equation? I'm getting lost with the factoring.
 
Dec 2015
972
128
Earth
\(\displaystyle 5\cos x -2(1-\cos^2 x )-1=0 \; \) ; \(\displaystyle \; 5\cos x -2 +2\cos^2 x -1 =2\cos^2 x +5\cos x -3=0 \).

\(\displaystyle \cos x = \dfrac{-5 \pm \sqrt{25-24}}{4}=\dfrac{-5 \pm 1 }{4}\).

\(\displaystyle x=\arccos \dfrac{-5 \pm 1 }{4} \).
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
$5\cos{x} - 2(1-\cos^2{x}) = 1$

$2\cos^2{x} + 5\cos{x} - 3 = 0$

let $u = \cos{x}$

$2u^2 + 5u - 3 = 0$

now factor the quadratic ...
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
\(\displaystyle 5\cos x -2(1-\cos^2 x )-1=0 \; \) ; \(\displaystyle \; 5\cos x -2 +2\cos^2 x -1 =2\cos^2 x +5\cos x -3=0 \).

\(\displaystyle \cos x = \dfrac{-5 \pm \sqrt{25-24}}{4}=\dfrac{-5 \pm 1 }{4}\).

\(\displaystyle x=\arccos \dfrac{-5 \pm 1 }{4} \).
\(\displaystyle \cos x = \dfrac{-5 \pm \sqrt{{\color{red}25+24}}}{4}=\dfrac{-5 \pm 7 }{4}\)

$\cos{x} = \dfrac{1}{2}$ is the only valid solution
 
Dec 2019
8
0
Windsor Ontario Canada
Thank you @skeeter, that's also what I ended up with; @idontknow had me scratching my head for a bit, but my solution matched up with the teacher's answer key. Thanks, guys, appreciate it.