Answer :
Certainly! Let's solve the given trigonometric equation step by step:
[tex]\[ 9 \cos^2 x - 4 \cos x - 1 = 0 \][/tex]
1. Substitute [tex]\(\cos(x)\)[/tex] with [tex]\(u\)[/tex]:
Let [tex]\(u = \cos(x)\)[/tex]. Then the equation becomes a quadratic in [tex]\(u\)[/tex]:
[tex]\[ 9u^2 - 4u - 1 = 0 \][/tex]
2. Solve the quadratic equation:
The general form of a quadratic equation is [tex]\(au^2 + bu + c = 0\)[/tex]. For our equation, [tex]\(a = 9\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -1\)[/tex].
3. Calculate the discriminant:
The discriminant of a quadratic equation [tex]\(au^2 + bu + c = 0\)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
Here, [tex]\( \Delta = (-4)^2 - 4 \cdot 9 \cdot (-1) = 16 + 36 = 52 \)[/tex].
4. Find the roots using the quadratic formula:
The quadratic formula states that the solutions for [tex]\(u\)[/tex] are given by:
[tex]\[ u = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Plugging in our values:
[tex]\[ u_{1} = \frac{-(-4) + \sqrt{52}}{2 \cdot 9} = \frac{4 + 2\sqrt{13}}{18} = \frac{2 + \sqrt{13}}{9} \approx 0.6228 \][/tex]
[tex]\[ u_{2} = \frac{-(-4) - \sqrt{52}}{2 \cdot 9} = \frac{4 - 2\sqrt{13}}{18} = \frac{2 - \sqrt{13}}{9} \approx -0.1784 \][/tex]
5. Convert [tex]\(u\)[/tex] back to [tex]\(x\)[/tex] using the arccosine function:
To find [tex]\(x\)[/tex], we take the arccosine of both [tex]\(u_{1}\)[/tex] and [tex]\(u_{2}\)[/tex]:
[tex]\[ x_{1} = \arccos(u_{1}) \approx \arccos(0.6228) \approx 0.8984 \, \text{(radians)} \][/tex]
[tex]\[ x_{2} = \arccos(u_{2}) \approx \arccos(-0.1784) \approx 1.7501 \, \text{(radians)} \][/tex]
6. Find additional solutions within the range [tex]\([0, 2\pi)\)[/tex]:
Since the cosine function is periodic and symmetric, we have additional solutions:
[tex]\[ x_{3} = 2\pi - x_{1} = 2\pi - 0.8984 \approx 5.3848 \, \text{(radians)} \][/tex]
[tex]\[ x_{4} = 2\pi - x_{2} = 2\pi - 1.7501 \approx 4.5330 \, \text{(radians)} \][/tex]
Hence, the detailed solutions for [tex]\(x\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\( x_{1} \approx 0.8984 \, \text{radians} \)[/tex]
- [tex]\( x_{2} \approx 1.7501 \, \text{radians} \)[/tex]
- [tex]\( x_{3} \approx 5.3848 \, \text{radians} \)[/tex]
- [tex]\( x_{4} \approx 4.5330 \, \text{radians} \)[/tex]
[tex]\[ 9 \cos^2 x - 4 \cos x - 1 = 0 \][/tex]
1. Substitute [tex]\(\cos(x)\)[/tex] with [tex]\(u\)[/tex]:
Let [tex]\(u = \cos(x)\)[/tex]. Then the equation becomes a quadratic in [tex]\(u\)[/tex]:
[tex]\[ 9u^2 - 4u - 1 = 0 \][/tex]
2. Solve the quadratic equation:
The general form of a quadratic equation is [tex]\(au^2 + bu + c = 0\)[/tex]. For our equation, [tex]\(a = 9\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -1\)[/tex].
3. Calculate the discriminant:
The discriminant of a quadratic equation [tex]\(au^2 + bu + c = 0\)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
Here, [tex]\( \Delta = (-4)^2 - 4 \cdot 9 \cdot (-1) = 16 + 36 = 52 \)[/tex].
4. Find the roots using the quadratic formula:
The quadratic formula states that the solutions for [tex]\(u\)[/tex] are given by:
[tex]\[ u = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Plugging in our values:
[tex]\[ u_{1} = \frac{-(-4) + \sqrt{52}}{2 \cdot 9} = \frac{4 + 2\sqrt{13}}{18} = \frac{2 + \sqrt{13}}{9} \approx 0.6228 \][/tex]
[tex]\[ u_{2} = \frac{-(-4) - \sqrt{52}}{2 \cdot 9} = \frac{4 - 2\sqrt{13}}{18} = \frac{2 - \sqrt{13}}{9} \approx -0.1784 \][/tex]
5. Convert [tex]\(u\)[/tex] back to [tex]\(x\)[/tex] using the arccosine function:
To find [tex]\(x\)[/tex], we take the arccosine of both [tex]\(u_{1}\)[/tex] and [tex]\(u_{2}\)[/tex]:
[tex]\[ x_{1} = \arccos(u_{1}) \approx \arccos(0.6228) \approx 0.8984 \, \text{(radians)} \][/tex]
[tex]\[ x_{2} = \arccos(u_{2}) \approx \arccos(-0.1784) \approx 1.7501 \, \text{(radians)} \][/tex]
6. Find additional solutions within the range [tex]\([0, 2\pi)\)[/tex]:
Since the cosine function is periodic and symmetric, we have additional solutions:
[tex]\[ x_{3} = 2\pi - x_{1} = 2\pi - 0.8984 \approx 5.3848 \, \text{(radians)} \][/tex]
[tex]\[ x_{4} = 2\pi - x_{2} = 2\pi - 1.7501 \approx 4.5330 \, \text{(radians)} \][/tex]
Hence, the detailed solutions for [tex]\(x\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\( x_{1} \approx 0.8984 \, \text{radians} \)[/tex]
- [tex]\( x_{2} \approx 1.7501 \, \text{radians} \)[/tex]
- [tex]\( x_{3} \approx 5.3848 \, \text{radians} \)[/tex]
- [tex]\( x_{4} \approx 4.5330 \, \text{radians} \)[/tex]
