Answer :
Certainly! Let's solve the equation [tex]\( 13 \sin^2{\theta} - 7 \sin{\theta} = 4 \)[/tex] algebraically. We need to find the values of [tex]\(\theta\)[/tex] in the interval [tex]\([0^\circ, 360^\circ)\)[/tex].
### Step 1: Substitution
Let's introduce a substitution to simplify the equation. Let [tex]\( x = \sin{\theta} \)[/tex]. Thus, the equation becomes:
[tex]\[ 13x^2 - 7x = 4 \][/tex]
### Step 2: Form a quadratic equation
Rearrange the equation into standard quadratic form:
[tex]\[ 13x^2 - 7x - 4 = 0 \][/tex]
### Step 3: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 13 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -4 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 13 \cdot (-4)}}{2 \cdot 13} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 + 208}}{26} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{257}}{26} \][/tex]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{7 + \sqrt{257}}{26} \][/tex]
[tex]\[ x_2 = \frac{7 - \sqrt{257}}{26} \][/tex]
### Step 4: Check the validity of solutions
We need to check that the solutions for [tex]\( x = \sin{\theta} \)[/tex] lie within the possible range of [tex]\(-1 \leq \sin{\theta} \leq 1\)[/tex].
1. [tex]\( x_1 = \frac{7 + \sqrt{257}}{26} \)[/tex]:
This value is greater than 1 (since [tex]\(\sqrt{257} \approx 16\)[/tex] and [tex]\(7 + 16 = 23\)[/tex], thus [tex]\(\frac{23}{26} < 1\)[/tex]).
2. [tex]\( x_2 = \frac{7 - \sqrt{257}}{26} \)[/tex]:
This value is definitely less than -1 since [tex]\(\sqrt{257} \approx 16\)[/tex] and [tex]\(7 - 16 = -9\)[/tex], thus [tex]\(\frac{-9}{26} > -1\)[/tex].
Since these values lie within the [tex]\(\sin{\theta}\)[/tex] range, we consider them for obtaining [tex]\(\theta\)[/tex].
### Step 5: Finding [tex]\(\theta\)[/tex]
Now, we solve for [tex]\(\theta\)[/tex] from [tex]\(\sin{\theta} = \frac{7 + \sqrt{257}}{26}\)[/tex].
For [tex]\(\sin{\theta} = x_1 = \frac{7 + \sqrt{257}}{26}\)[/tex]:
[tex]\[ \theta = \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \][/tex]
- This corresponds to:
[tex]\[ \theta = \arcsin\left(x_1\right) \text{ and } \theta = 180^\circ - \arcsin\left(x_1\right) \quad (\text{since sine is positive in both quadrants I and II}) \][/tex]
For [tex]\(\sin{\theta} = x_2 = \frac{7 - \sqrt{257}}{26}\)[/tex]:
[tex]\[ \theta = \arcsin\left(x_2\right) \][/tex]
- This corresponds to:
[tex]\[ \theta = \arcsin\left(x_2\right) \text{ and } \theta = 180^\circ - \arcsin\left(x_2\right) \quad (\text{since sine is positive in both quadrants I and II}) \][/tex]
### Step 6: List all the solutions
The solutions for [tex]\(\theta\)[/tex] in the range [tex]\([0^\circ, 360^\circ)\)[/tex] are:
- [tex]\(\theta = \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \approx 195.885^\circ\)[/tex]
- [tex]\(\theta = 180^\circ - \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \approx -63.855^\circ \text{ (adjust for positive angle: 360^\circ - 63.855^\circ = 296.145^\circ)}\)[/tex]
- [tex]\(\theta = \arcsin\left(\frac{7 - \sqrt{257}}{26}\right) \approx 369.602^\circ \text{ (exclude as it's out of range)}\)[/tex]
- [tex]\(\theta = 180^\circ - \arcsin\left(\frac{7 - \sqrt{257}}{26}\right) \approx 629.341^\circ \text{ (exclude as it's out of range)}\)[/tex]
After checking the feasible solutions, we have:
[tex]\[ \theta \approx 195.885^\circ, 296.145^\circ \][/tex]
These are the valid solutions in the interval [tex]\([0^\circ, 360^\circ)\)[/tex].
### Step 1: Substitution
Let's introduce a substitution to simplify the equation. Let [tex]\( x = \sin{\theta} \)[/tex]. Thus, the equation becomes:
[tex]\[ 13x^2 - 7x = 4 \][/tex]
### Step 2: Form a quadratic equation
Rearrange the equation into standard quadratic form:
[tex]\[ 13x^2 - 7x - 4 = 0 \][/tex]
### Step 3: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 13 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -4 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 13 \cdot (-4)}}{2 \cdot 13} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 + 208}}{26} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{257}}{26} \][/tex]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{7 + \sqrt{257}}{26} \][/tex]
[tex]\[ x_2 = \frac{7 - \sqrt{257}}{26} \][/tex]
### Step 4: Check the validity of solutions
We need to check that the solutions for [tex]\( x = \sin{\theta} \)[/tex] lie within the possible range of [tex]\(-1 \leq \sin{\theta} \leq 1\)[/tex].
1. [tex]\( x_1 = \frac{7 + \sqrt{257}}{26} \)[/tex]:
This value is greater than 1 (since [tex]\(\sqrt{257} \approx 16\)[/tex] and [tex]\(7 + 16 = 23\)[/tex], thus [tex]\(\frac{23}{26} < 1\)[/tex]).
2. [tex]\( x_2 = \frac{7 - \sqrt{257}}{26} \)[/tex]:
This value is definitely less than -1 since [tex]\(\sqrt{257} \approx 16\)[/tex] and [tex]\(7 - 16 = -9\)[/tex], thus [tex]\(\frac{-9}{26} > -1\)[/tex].
Since these values lie within the [tex]\(\sin{\theta}\)[/tex] range, we consider them for obtaining [tex]\(\theta\)[/tex].
### Step 5: Finding [tex]\(\theta\)[/tex]
Now, we solve for [tex]\(\theta\)[/tex] from [tex]\(\sin{\theta} = \frac{7 + \sqrt{257}}{26}\)[/tex].
For [tex]\(\sin{\theta} = x_1 = \frac{7 + \sqrt{257}}{26}\)[/tex]:
[tex]\[ \theta = \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \][/tex]
- This corresponds to:
[tex]\[ \theta = \arcsin\left(x_1\right) \text{ and } \theta = 180^\circ - \arcsin\left(x_1\right) \quad (\text{since sine is positive in both quadrants I and II}) \][/tex]
For [tex]\(\sin{\theta} = x_2 = \frac{7 - \sqrt{257}}{26}\)[/tex]:
[tex]\[ \theta = \arcsin\left(x_2\right) \][/tex]
- This corresponds to:
[tex]\[ \theta = \arcsin\left(x_2\right) \text{ and } \theta = 180^\circ - \arcsin\left(x_2\right) \quad (\text{since sine is positive in both quadrants I and II}) \][/tex]
### Step 6: List all the solutions
The solutions for [tex]\(\theta\)[/tex] in the range [tex]\([0^\circ, 360^\circ)\)[/tex] are:
- [tex]\(\theta = \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \approx 195.885^\circ\)[/tex]
- [tex]\(\theta = 180^\circ - \arcsin\left(\frac{7 + \sqrt{257}}{26}\right) \approx -63.855^\circ \text{ (adjust for positive angle: 360^\circ - 63.855^\circ = 296.145^\circ)}\)[/tex]
- [tex]\(\theta = \arcsin\left(\frac{7 - \sqrt{257}}{26}\right) \approx 369.602^\circ \text{ (exclude as it's out of range)}\)[/tex]
- [tex]\(\theta = 180^\circ - \arcsin\left(\frac{7 - \sqrt{257}}{26}\right) \approx 629.341^\circ \text{ (exclude as it's out of range)}\)[/tex]
After checking the feasible solutions, we have:
[tex]\[ \theta \approx 195.885^\circ, 296.145^\circ \][/tex]
These are the valid solutions in the interval [tex]\([0^\circ, 360^\circ)\)[/tex].
