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To solve the equation [tex]\(\cos(2\theta) + 18\sin^2(\theta) = 13\)[/tex] in the interval [tex]\(0 \leq \theta < 2\pi\)[/tex], let’s follow these steps:
### Step 1: Recall trigonometric identities
We know that:
[tex]\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \][/tex]
### Step 2: Substitute the identity into the equation
Substitute [tex]\(\cos(2\theta)\)[/tex] with [tex]\(1 - 2\sin^2(\theta)\)[/tex] in the given equation:
[tex]\[ 1 - 2\sin^2(\theta) + 18\sin^2(\theta) = 13 \][/tex]
### Step 3: Simplify the equation
Combine like terms involving [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ 1 + 16\sin^2(\theta) = 13 \][/tex]
### Step 4: Isolate the trigonometric function
Subtract 1 from both sides of the equation:
[tex]\[ 16\sin^2(\theta) = 12 \][/tex]
### Step 5: Solve for [tex]\(\sin^2(\theta)\)[/tex]
Divide both sides by 16:
[tex]\[ \sin^2(\theta) = \frac{12}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{3}{4} \][/tex]
### Step 6: Solve for [tex]\(\sin(\theta)\)[/tex]
Take the square root of both sides (consider both positive and negative roots):
[tex]\[ \sin(\theta) = \pm\sqrt{\frac{3}{4}} \][/tex]
[tex]\[ \sin(\theta) = \pm\frac{\sqrt{3}}{2} \][/tex]
### Step 7: Find [tex]\(\theta\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex]
There are four angles within the interval [tex]\(0 \leq \theta < 2\pi\)[/tex] where [tex]\(\sin(\theta) = \pm\frac{\sqrt{3}}{2}\)[/tex]:
1. [tex]\(\theta = \frac{\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex])
2. [tex]\(\theta = \frac{2\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex])
3. [tex]\(\theta = \frac{4\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)[/tex])
4. [tex]\(\theta = \frac{5\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)[/tex])
### Conclusion
The solutions to the equation [tex]\(\cos(2\theta) + 18\sin^2(\theta) = 13\)[/tex] in the interval [tex]\(0 \leq \theta < 2\pi\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \][/tex]
### Step 1: Recall trigonometric identities
We know that:
[tex]\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \][/tex]
### Step 2: Substitute the identity into the equation
Substitute [tex]\(\cos(2\theta)\)[/tex] with [tex]\(1 - 2\sin^2(\theta)\)[/tex] in the given equation:
[tex]\[ 1 - 2\sin^2(\theta) + 18\sin^2(\theta) = 13 \][/tex]
### Step 3: Simplify the equation
Combine like terms involving [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ 1 + 16\sin^2(\theta) = 13 \][/tex]
### Step 4: Isolate the trigonometric function
Subtract 1 from both sides of the equation:
[tex]\[ 16\sin^2(\theta) = 12 \][/tex]
### Step 5: Solve for [tex]\(\sin^2(\theta)\)[/tex]
Divide both sides by 16:
[tex]\[ \sin^2(\theta) = \frac{12}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{3}{4} \][/tex]
### Step 6: Solve for [tex]\(\sin(\theta)\)[/tex]
Take the square root of both sides (consider both positive and negative roots):
[tex]\[ \sin(\theta) = \pm\sqrt{\frac{3}{4}} \][/tex]
[tex]\[ \sin(\theta) = \pm\frac{\sqrt{3}}{2} \][/tex]
### Step 7: Find [tex]\(\theta\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex]
There are four angles within the interval [tex]\(0 \leq \theta < 2\pi\)[/tex] where [tex]\(\sin(\theta) = \pm\frac{\sqrt{3}}{2}\)[/tex]:
1. [tex]\(\theta = \frac{\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex])
2. [tex]\(\theta = \frac{2\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex])
3. [tex]\(\theta = \frac{4\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)[/tex])
4. [tex]\(\theta = \frac{5\pi}{3}\)[/tex] (since [tex]\(\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)[/tex])
### Conclusion
The solutions to the equation [tex]\(\cos(2\theta) + 18\sin^2(\theta) = 13\)[/tex] in the interval [tex]\(0 \leq \theta < 2\pi\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \][/tex]
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