Solve the equation on the interval [0,2𝜋). NO CALCULATOR.

Question

Solve the equation on the interval [0,2𝜋).

2cos(^2)𝑥 + 3cos𝑥+1 = 0

Answer 1

2cos²(x) + 3cos(x) + 1 = 0.

Let's think of another way to look at this. This looks very similar to a quadratic function, of the form

ax²+bx+c=0. If we let a dummy variable, say u = cos(x), then we can rewrite this as

2u²+3u+1=0

This becomes a quadratic that we know how to solve, using the quadratic formula

a=2, b=3, c=1, -3±√(3)^2-4*(2)*(1) / 2(2) = -3±√(9-8)/4 = -3±√(1)/4 = (-3±1)/4

u₁=(-3+1)/4=-2/4=-1/2

u₂=(-3-1)/4=(-4)/4=-1

u₁=-1/2

u₂=-1

So now that we have our solutions of this, we have to solve them in the context of [0, 2π).

First, u₁: Solve cos(x)=-1/2, 0≤x<2π

Take the arccos to get x=arccos(-1/2)=2π/3 and 4π/3. We do not need to include additional periods here, as we within our domain here and any multiple of 2π will take us out.

Now, u₂: Solve cos(x)=-1, 0≤x<2π

x=arccos(-1)=π. This is the only solution which stays within our bounds.

So, finally, we just now list all of our solutions, which are

x=2π/3, x=π, x=4π/3.

I hope this is clear and able to help you out! Good luck!

Answer 2

Solve the equation on the interval [0, 2𝜋). NO CALCULATOR. 2 cos² 𝑥 + 3 cos 𝑥 + 1 = 0

Detailed Solution:

2 cos² 𝑥 + 3 cos 𝑥 + 1 = 0 [0, 2𝜋)

2 cos² 𝑥 + 3 cos 𝑥 + 1 = 0 ==> (cos 𝑥 + 1) (2 cos 𝑥 + 1) = 0 ==> cos 𝑥 + 1 = 0 and 2 cos 𝑥 + 1 = 0

cos 𝑥 + 1 = 0 ==> cos 𝑥 = −1 ==> 𝑥 = cos^-1 (−1) ==> 𝑥 = 𝜋

2 cos 𝑥 + 1 = 0 ==> 2 cos 𝑥 = −1 ==> cos 𝑥 = −1/2 (Q 2 & Q3)

Quadrant 2: cos 𝑥 = −1/2 ==> 𝑥 = cos^-1 (−1/2) ==> 𝑥 = 𝜋 − 𝜋/3 ==> 𝑥 = 2𝜋/3

Quadrant 3: cos 𝑥 = −1/2 ==> 𝑥 = cos^-1 (−1/2) ==> 𝑥 = 𝜋 + 𝜋/3 ==> 𝑥 = 4𝜋/3

Therefore: x = {2𝜋/3, 𝜋, 4𝜋/3}

Answer 3

Factor using Algebraic techniques.

Use zero-product rule