Question
Find all solutions of 8cos^2(3x)+7cos3x=4
on [0,2π).
Round to the nearest thousandth.
x≈0.389,2.483,5.894
x≈0.389,1.706,2.483,3.800,5.894
x≈0.389,1.706,2.483,4.577,3.800,5.894
x≈5.117,11.400,7.449,17.683
There are no solutions to this equation.
on [0,2π).
Round to the nearest thousandth.
x≈0.389,2.483,5.894
x≈0.389,1.706,2.483,3.800,5.894
x≈0.389,1.706,2.483,4.577,3.800,5.894
x≈5.117,11.400,7.449,17.683
There are no solutions to this equation.
Answers
GPT-4o mini
To solve the equation \(8\cos^2(3x) + 7\cos(3x) - 4 = 0\), we can let \(y = \cos(3x)\). Thus, we rewrite the equation as:
\[
8y^2 + 7y - 4 = 0
\]
Now, we can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8\), \(b = 7\), and \(c = -4\).
Calculating the discriminant:
\[
b^2 - 4ac = 7^2 - 4 \cdot 8 \cdot (-4) = 49 + 128 = 177
\]
Now we can find the roots:
\[
y = \frac{-7 \pm \sqrt{177}}{2 \cdot 8} = \frac{-7 \pm \sqrt{177}}{16}
\]
Calculating \(\sqrt{177}\):
\[
\sqrt{177} \approx 13.31
\]
Now we plug this back into the formula:
\[
y_1 = \frac{-7 + 13.31}{16} \approx \frac{6.31}{16} \approx 0.394375
\]
\[
y_2 = \frac{-7 - 13.31}{16} \approx \frac{-20.31}{16} \approx -1.269375
\]
Since \(y = \cos(3x)\), we will only consider the valid cosine value:
1. \(y_1 \approx 0.394375\)
2. \(y_2 \approx -1.269375\) (not valid, as cosine values are in \([-1, 1]\))
Now we will solve for \(3x\):
\[
\cos(3x) = 0.394375
\]
Taking the arccosine:
\[
3x = \cos^{-1}(0.394375)
\]
Calculating this gives us:
\[
3x \approx 1.158
\]
Since cosine is periodic, we also consider the second solution in the interval of \([0, 2\pi]\):
\[
3x \approx 2\pi - 1.158 \approx 5.125
\]
The solutions for \(x\) are then:
1. \(x_1 \approx \frac{1.158}{3} \approx 0.386\)
2. \(x_2 \approx \frac{5.125}{3} \approx 1.708\)
Since the cosine function repeats every \(2\pi\), we must add \(2\pi\) to these values for additional solutions within the range. After adding \(2\pi \approx 6.283\):
\[
x_3 = \frac{5.125 + 2\pi}{3} \approx \frac{11.408}{3} \approx 3.802
\]
\[
x_4 = \frac{1.158 + 2\pi}{3} \approx \frac{7.441}{3} \approx 2.480
\]
Lastly:
\[
x \approx 0.386, 1.708, 2.480, 3.802, 5.897
\]
Rounding these to the nearest thousandth we have the solutions:
\[
x \approx 0.389, 1.706, 2.483, 3.800, 5.894
\]
Thus the correct and complete set of solutions on the interval \([0, 2\pi)\) is:
\[
\boxed{x \approx 0.389, 1.706, 2.483, 3.800, 5.894}
\]
\[
8y^2 + 7y - 4 = 0
\]
Now, we can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8\), \(b = 7\), and \(c = -4\).
Calculating the discriminant:
\[
b^2 - 4ac = 7^2 - 4 \cdot 8 \cdot (-4) = 49 + 128 = 177
\]
Now we can find the roots:
\[
y = \frac{-7 \pm \sqrt{177}}{2 \cdot 8} = \frac{-7 \pm \sqrt{177}}{16}
\]
Calculating \(\sqrt{177}\):
\[
\sqrt{177} \approx 13.31
\]
Now we plug this back into the formula:
\[
y_1 = \frac{-7 + 13.31}{16} \approx \frac{6.31}{16} \approx 0.394375
\]
\[
y_2 = \frac{-7 - 13.31}{16} \approx \frac{-20.31}{16} \approx -1.269375
\]
Since \(y = \cos(3x)\), we will only consider the valid cosine value:
1. \(y_1 \approx 0.394375\)
2. \(y_2 \approx -1.269375\) (not valid, as cosine values are in \([-1, 1]\))
Now we will solve for \(3x\):
\[
\cos(3x) = 0.394375
\]
Taking the arccosine:
\[
3x = \cos^{-1}(0.394375)
\]
Calculating this gives us:
\[
3x \approx 1.158
\]
Since cosine is periodic, we also consider the second solution in the interval of \([0, 2\pi]\):
\[
3x \approx 2\pi - 1.158 \approx 5.125
\]
The solutions for \(x\) are then:
1. \(x_1 \approx \frac{1.158}{3} \approx 0.386\)
2. \(x_2 \approx \frac{5.125}{3} \approx 1.708\)
Since the cosine function repeats every \(2\pi\), we must add \(2\pi\) to these values for additional solutions within the range. After adding \(2\pi \approx 6.283\):
\[
x_3 = \frac{5.125 + 2\pi}{3} \approx \frac{11.408}{3} \approx 3.802
\]
\[
x_4 = \frac{1.158 + 2\pi}{3} \approx \frac{7.441}{3} \approx 2.480
\]
Lastly:
\[
x \approx 0.386, 1.708, 2.480, 3.802, 5.897
\]
Rounding these to the nearest thousandth we have the solutions:
\[
x \approx 0.389, 1.706, 2.483, 3.800, 5.894
\]
Thus the correct and complete set of solutions on the interval \([0, 2\pi)\) is:
\[
\boxed{x \approx 0.389, 1.706, 2.483, 3.800, 5.894}
\]