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

#### JamesM0025

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

#### skeeter

Math Team
change $\sin^2{x}$ to $1-\cos^2{x}$ and solve the resulting quadratic equation for $\cos{x}$

• JamesM0025, topsquark and idontknow

#### JamesM0025

Do you mind writing all the steps for the equation? I'm getting lost with the factoring.

#### idontknow

$$\displaystyle 5\cos x -2(1-\cos^2 x )-1=0 \;$$ ; $$\displaystyle \; 5\cos x -2 +2\cos^2 x -1 =2\cos^2 x +5\cos x -3=0$$.

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

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

• topsquark and JamesM0025

#### skeeter

Math Team
$5\cos{x} - 2(1-\cos^2{x}) = 1$

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

let $u = \cos{x}$

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

• topsquark and JamesM0025

#### skeeter

Math Team
$$\displaystyle 5\cos x -2(1-\cos^2 x )-1=0 \;$$ ; $$\displaystyle \; 5\cos x -2 +2\cos^2 x -1 =2\cos^2 x +5\cos x -3=0$$.

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

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

$\cos{x} = \dfrac{1}{2}$ is the only valid solution

• topsquark and JamesM0025

#### JamesM0025

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.