Question
Find all solutions of 2(cos3x)^2+6cos3x=3 on [0,2π).
Round to the nearest thousandth.
x≈0.373,2.467,5.910
x≈0.373,1.721,2.467,3.816,5.910
x≈0.373,1.721,2.467,4.562,3.816,5.910
x≈5.164,11.447,7.402,17.730
There are no solutions to this equation.
Round to the nearest thousandth.
x≈0.373,2.467,5.910
x≈0.373,1.721,2.467,3.816,5.910
x≈0.373,1.721,2.467,4.562,3.816,5.910
x≈5.164,11.447,7.402,17.730
There are no solutions to this equation.
Answers
GPT-4o mini
To solve the equation \( 2(\cos(3x))^2 + 6\cos(3x) - 3 = 0 \), we can start by letting \( y = \cos(3x) \). This transforms the equation into a standard quadratic equation:
\[
2y^2 + 6y - 3 = 0
\]
Next, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b = 6 \), and \( c = -3 \):
\[
y = \frac{-6 \pm \sqrt{6^2 - 4(2)(-3)}}{2(2)}
\]
Calculating the discriminant:
\[
6^2 - 4 \cdot 2 \cdot (-3) = 36 + 24 = 60
\]
Now substituting back into the formula:
\[
y = \frac{-6 \pm \sqrt{60}}{4}
\]
Simplifying \( \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \):
\[
y = \frac{-6 \pm 2\sqrt{15}}{4} = \frac{-3 \pm \sqrt{15}}{2}
\]
This gives us two potential values for \( y \):
\[
y_1 = \frac{-3 + \sqrt{15}}{2}, \quad y_2 = \frac{-3 - \sqrt{15}}{2}
\]
Next, we need to evaluate these values to see if they are in the range of the cosine function, which is \([-1, 1]\).
Calculating \( y_1 \):
\[
\sqrt{15} \approx 3.873
\]
\[
y_1 \approx \frac{-3 + 3.873}{2} \approx \frac{0.873}{2} \approx 0.4365 \quad (\text{valid since } 0.4365 \in [-1, 1])
\]
Calculating \( y_2 \):
\[
y_2 \approx \frac{-3 - 3.873}{2} \approx \frac{-6.873}{2} \approx -3.4365 \quad (\text{not valid since } -3.4365 \notin [-1, 1])
\]
Now we only consider \( y_1 \):
\[
\cos(3x) = \frac{-3 + \sqrt{15}}{2}
\]
To find \( x \), we have:
\[
3x = \cos^{-1}\left(\frac{-3 + \sqrt{15}}{2}\right)
\]
This gives us two solutions from the cosine function:
1. \( 3x = \cos^{-1}(y_1) \)
2. \( 3x = 2\pi - \cos^{-1}(y_1) \)
Calculating \( \cos^{-1}(y_1) \):
Using a calculator:
- \( \cos^{-1}(0.4365) \approx 1.129 \)
Then, we can find the angles:
1. \( 3x_1 = 1.129 \implies x_1 = \frac{1.129}{3} \approx 0.376 \)
2. \( 3x_2 = 2\pi - 1.129 \implies x_2 = \frac{2\pi - 1.129}{3} \)
Calculating \( x_2 \):
\[
2\pi \approx 6.283 \implies 2\pi - 1.129 \approx 5.154 \implies x_2 \approx \frac{5.154}{3} \approx 1.718
\]
Now, accounting for the periodicity of cosine, the next solutions can be derived by adding multiples of \( \frac{2\pi}{3} \):
\[
x_3 = x_1 + \frac{2\pi}{3} \approx 0.376 + \frac{2\pi}{3} \approx 0.376 + 2.094 = 2.470
\]
\[
x_4 = x_2 + \frac{2\pi}{3} \approx 1.718 + 2.094 \approx 3.812
\]
Now check if the next cycle fits in the range \( [0, 2\pi) \):
\[
x_5 = x_1 + 4\pi/3 \approx 0.376 + 4.188 = 4.564
\]
\[
x_6 = x_2 + 4\pi/3 \approx 1.718 + 4.188 = 5.906
\]
Collecting all solutions in range \( [0, 2\pi) \):
\[
x \approx 0.376, \quad 1.718, \quad 2.470, \quad 4.564, \quad 5.906
\]
Rounding to the nearest thousandths, the solutions are:
\[
\boxed{x \approx 0.376, 1.718, 2.470, 4.564, 5.906}
\]
\[
2y^2 + 6y - 3 = 0
\]
Next, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b = 6 \), and \( c = -3 \):
\[
y = \frac{-6 \pm \sqrt{6^2 - 4(2)(-3)}}{2(2)}
\]
Calculating the discriminant:
\[
6^2 - 4 \cdot 2 \cdot (-3) = 36 + 24 = 60
\]
Now substituting back into the formula:
\[
y = \frac{-6 \pm \sqrt{60}}{4}
\]
Simplifying \( \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \):
\[
y = \frac{-6 \pm 2\sqrt{15}}{4} = \frac{-3 \pm \sqrt{15}}{2}
\]
This gives us two potential values for \( y \):
\[
y_1 = \frac{-3 + \sqrt{15}}{2}, \quad y_2 = \frac{-3 - \sqrt{15}}{2}
\]
Next, we need to evaluate these values to see if they are in the range of the cosine function, which is \([-1, 1]\).
Calculating \( y_1 \):
\[
\sqrt{15} \approx 3.873
\]
\[
y_1 \approx \frac{-3 + 3.873}{2} \approx \frac{0.873}{2} \approx 0.4365 \quad (\text{valid since } 0.4365 \in [-1, 1])
\]
Calculating \( y_2 \):
\[
y_2 \approx \frac{-3 - 3.873}{2} \approx \frac{-6.873}{2} \approx -3.4365 \quad (\text{not valid since } -3.4365 \notin [-1, 1])
\]
Now we only consider \( y_1 \):
\[
\cos(3x) = \frac{-3 + \sqrt{15}}{2}
\]
To find \( x \), we have:
\[
3x = \cos^{-1}\left(\frac{-3 + \sqrt{15}}{2}\right)
\]
This gives us two solutions from the cosine function:
1. \( 3x = \cos^{-1}(y_1) \)
2. \( 3x = 2\pi - \cos^{-1}(y_1) \)
Calculating \( \cos^{-1}(y_1) \):
Using a calculator:
- \( \cos^{-1}(0.4365) \approx 1.129 \)
Then, we can find the angles:
1. \( 3x_1 = 1.129 \implies x_1 = \frac{1.129}{3} \approx 0.376 \)
2. \( 3x_2 = 2\pi - 1.129 \implies x_2 = \frac{2\pi - 1.129}{3} \)
Calculating \( x_2 \):
\[
2\pi \approx 6.283 \implies 2\pi - 1.129 \approx 5.154 \implies x_2 \approx \frac{5.154}{3} \approx 1.718
\]
Now, accounting for the periodicity of cosine, the next solutions can be derived by adding multiples of \( \frac{2\pi}{3} \):
\[
x_3 = x_1 + \frac{2\pi}{3} \approx 0.376 + \frac{2\pi}{3} \approx 0.376 + 2.094 = 2.470
\]
\[
x_4 = x_2 + \frac{2\pi}{3} \approx 1.718 + 2.094 \approx 3.812
\]
Now check if the next cycle fits in the range \( [0, 2\pi) \):
\[
x_5 = x_1 + 4\pi/3 \approx 0.376 + 4.188 = 4.564
\]
\[
x_6 = x_2 + 4\pi/3 \approx 1.718 + 4.188 = 5.906
\]
Collecting all solutions in range \( [0, 2\pi) \):
\[
x \approx 0.376, \quad 1.718, \quad 2.470, \quad 4.564, \quad 5.906
\]
Rounding to the nearest thousandths, the solutions are:
\[
\boxed{x \approx 0.376, 1.718, 2.470, 4.564, 5.906}
\]