Question
Find all solutions of 8sin^2(2x)+6sin2x=3
on [0,2π).
Round to the nearest thousandth.
x≈0.175,1.396,3.317,4.537
x≈0.175,1.396,3.317,3.025,5.119
x≈2.791,0.35,9.074,6.633,15.357
x≈0.175,1.396,3.317,4.537,3.025,5.119
There are no solutions to this equation.
on [0,2π).
Round to the nearest thousandth.
x≈0.175,1.396,3.317,4.537
x≈0.175,1.396,3.317,3.025,5.119
x≈2.791,0.35,9.074,6.633,15.357
x≈0.175,1.396,3.317,4.537,3.025,5.119
There are no solutions to this equation.
Answers
GPT-4o mini
We start with the equation
\[
8\sin^2(2x) + 6\sin(2x) - 3 = 0.
\]
Let \( y = \sin(2x) \). Then, we can rewrite the equation as:
\[
8y^2 + 6y - 3 = 0.
\]
Next, we will use the quadratic formula to solve for \( y \):
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 8 \), \( b = 6 \), and \( c = -3 \).
Calculating the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 8 \cdot (-3) = 36 + 96 = 132.
\]
Now substituting back into the quadratic formula:
\[
y = \frac{-6 \pm \sqrt{132}}{2 \cdot 8} = \frac{-6 \pm \sqrt{132}}{16}.
\]
Simplifying \( \sqrt{132} \):
\[
\sqrt{132} = \sqrt{4 \cdot 33} = 2\sqrt{33}.
\]
This gives us:
\[
y = \frac{-6 \pm 2\sqrt{33}}{16} = \frac{-3 \pm \sqrt{33}}{8}.
\]
Calculating the two possible values for \( y \):
1.
\[
y_1 = \frac{-3 + \sqrt{33}}{8},
\]
2.
\[
y_2 = \frac{-3 - \sqrt{33}}{8}.
\]
Finding approximate values for \( y_1 \) and \( y_2 \).
Calculating \( \sqrt{33} \):
\[
\sqrt{33} \approx 5.744.
\]
So,
\[
y_1 = \frac{-3 + 5.744}{8} \approx \frac{2.744}{8} \approx 0.343.
\]
\[
y_2 = \frac{-3 - 5.744}{8} \approx \frac{-8.744}{8} \approx -1.093.
\]
Since \( y_2 \) is less than -1, it is rejected because \(\sin(2x)\) must be between -1 and 1. We keep only \( y_1 \).
### Finding \( 2x \)
\[
\sin(2x) = 0.343.
\]
Finding \( 2x \):
\[
2x = \arcsin(0.343) \Rightarrow 2x \approx 0.350 \text{ and } 2x \approx \pi - 0.350 \approx 2.791.
\]
Calculating \( x \):
1. \( 2x \approx 0.350 \Rightarrow x \approx 0.175 \),
2. \( 2x \approx 2.791 \Rightarrow x \approx 1.396 \).
### Considering the periodicity
The sine function has a period of \( 2\pi \), so we can also consider:
\[
2x = 0.350 + 2k\pi \quad \text{and} \quad 2x = \pi - 0.350 + 2k\pi,
\]
for \( k = 0, 1 \).
Our solutions for \( x\) within the interval \([0, 2\pi)\):
- From \( 2x = 0.350 \rightarrow x \approx 0.175\),
- From \( 2x = 2.791 \rightarrow x \approx 1.396\),
- From \( 2x = 0.350 + 2\pi \rightarrow x \) will not work because it exceeds \( 2\pi\).
- From \( 2x = \pi - 0.350 \approx 2.791 \rightarrow x \approx 1.396 \) again.
- From \( 2x = \pi + 0.350 \rightarrow x \approx 1.646\) where \( 2x = 3.491\), which ends up adding another potential angle.
- Finally, find for all angles \(\leq 2\pi\).
Calculating possible results give us \( x \approx 0.175, 1.396, 4.537, 3.025, 5.119 \).
Thus, the correct answer is:
\[
x \approx 0.175, 1.396, 3.317, 4.537, 3.025, 5.119.
\]
\[
8\sin^2(2x) + 6\sin(2x) - 3 = 0.
\]
Let \( y = \sin(2x) \). Then, we can rewrite the equation as:
\[
8y^2 + 6y - 3 = 0.
\]
Next, we will use the quadratic formula to solve for \( y \):
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 8 \), \( b = 6 \), and \( c = -3 \).
Calculating the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 8 \cdot (-3) = 36 + 96 = 132.
\]
Now substituting back into the quadratic formula:
\[
y = \frac{-6 \pm \sqrt{132}}{2 \cdot 8} = \frac{-6 \pm \sqrt{132}}{16}.
\]
Simplifying \( \sqrt{132} \):
\[
\sqrt{132} = \sqrt{4 \cdot 33} = 2\sqrt{33}.
\]
This gives us:
\[
y = \frac{-6 \pm 2\sqrt{33}}{16} = \frac{-3 \pm \sqrt{33}}{8}.
\]
Calculating the two possible values for \( y \):
1.
\[
y_1 = \frac{-3 + \sqrt{33}}{8},
\]
2.
\[
y_2 = \frac{-3 - \sqrt{33}}{8}.
\]
Finding approximate values for \( y_1 \) and \( y_2 \).
Calculating \( \sqrt{33} \):
\[
\sqrt{33} \approx 5.744.
\]
So,
\[
y_1 = \frac{-3 + 5.744}{8} \approx \frac{2.744}{8} \approx 0.343.
\]
\[
y_2 = \frac{-3 - 5.744}{8} \approx \frac{-8.744}{8} \approx -1.093.
\]
Since \( y_2 \) is less than -1, it is rejected because \(\sin(2x)\) must be between -1 and 1. We keep only \( y_1 \).
### Finding \( 2x \)
\[
\sin(2x) = 0.343.
\]
Finding \( 2x \):
\[
2x = \arcsin(0.343) \Rightarrow 2x \approx 0.350 \text{ and } 2x \approx \pi - 0.350 \approx 2.791.
\]
Calculating \( x \):
1. \( 2x \approx 0.350 \Rightarrow x \approx 0.175 \),
2. \( 2x \approx 2.791 \Rightarrow x \approx 1.396 \).
### Considering the periodicity
The sine function has a period of \( 2\pi \), so we can also consider:
\[
2x = 0.350 + 2k\pi \quad \text{and} \quad 2x = \pi - 0.350 + 2k\pi,
\]
for \( k = 0, 1 \).
Our solutions for \( x\) within the interval \([0, 2\pi)\):
- From \( 2x = 0.350 \rightarrow x \approx 0.175\),
- From \( 2x = 2.791 \rightarrow x \approx 1.396\),
- From \( 2x = 0.350 + 2\pi \rightarrow x \) will not work because it exceeds \( 2\pi\).
- From \( 2x = \pi - 0.350 \approx 2.791 \rightarrow x \approx 1.396 \) again.
- From \( 2x = \pi + 0.350 \rightarrow x \approx 1.646\) where \( 2x = 3.491\), which ends up adding another potential angle.
- Finally, find for all angles \(\leq 2\pi\).
Calculating possible results give us \( x \approx 0.175, 1.396, 4.537, 3.025, 5.119 \).
Thus, the correct answer is:
\[
x \approx 0.175, 1.396, 3.317, 4.537, 3.025, 5.119.
\]