Answer :
Certainly! Let's solve the equation [tex]\( \cos \left(2x + \frac{\pi}{3}\right) = \cos \frac{\pi}{5} \)[/tex] given the constraint [tex]\( 10^\circ \leqslant x \leqslant 2\pi \)[/tex].
We begin by using the property of the cosine function that states if [tex]\( \cos A = \cos B \)[/tex], then [tex]\( A = 2k\pi \pm B \)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex].
For our equation, we set:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi \pm \frac{\pi}{5} \][/tex]
Let's first solve using the positive case:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi + \frac{\pi}{5} \][/tex]
Reorganize to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 2k\pi + \frac{\pi}{5} - \frac{\pi}{3} \][/tex]
[tex]\[ 2x = 2k\pi + \frac{\pi}{15} \left(5 - 3\right) \][/tex]
[tex]\[ 2x = 2k\pi + \frac{2\pi}{15} \][/tex]
[tex]\[ x = k\pi + \frac{\pi}{15} \][/tex]
Now, let's solve using the negative case:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi - \frac{\pi}{5} \][/tex]
Reorganize to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 2k\pi - \frac{\pi}{5} - \frac{\pi}{3} \][/tex]
[tex]\[ 2x = 2k\pi - \frac{\pi}{15} \left(5 + 3\right) \][/tex]
[tex]\[ 2x = 2k\pi - \frac{8\pi}{15} \][/tex]
[tex]\[ x = k\pi - \frac{4\pi}{15} \][/tex]
Next, we consider the constraint [tex]\( 10^\circ \leqslant x \leqslant 2\pi \)[/tex], which translates [tex]\( 10^\circ \)[/tex] to radians:
[tex]\[ 10^\circ = \frac{10\pi}{180} = \frac{\pi}{18} \][/tex]
The constraint becomes:
[tex]\[ \frac{\pi}{18} \leq x \leq 2\pi \][/tex]
We now solve for [tex]\( x \)[/tex] within this range.
For [tex]\( x = k\pi + \frac{\pi}{15} \)[/tex] we check if the possible [tex]\( x \)[/tex] values meet the constraint:
1. [tex]\( k = 0 \)[/tex]:
[tex]\[ x = \frac{\pi}{15} \][/tex]
[tex]\[ \frac{\pi}{15} \approx 0.209 \quad \text{radians (which is less than } \frac{\pi}{18}\text{), so it's not in the valid range.} \][/tex]
2. [tex]\( k = 1 \)[/tex]:
[tex]\[ x = \pi + \frac{\pi}{15} = \frac{16\pi}{15} \][/tex]
[tex]\[ \frac{16\pi}{15} \approx 3.351 \quad \text{radians (which is greater than } \frac{\pi}{18} \text{ and less than } 2\pi\text{, so it meets the condition.} \][/tex]
For [tex]\( x = k\pi - \frac{4\pi}{15} \)[/tex], check the possible [tex]\( x \)[/tex] values:
1. [tex]\( k = 0 \)[/tex]:
[tex]\[ x = -\frac{4\pi}{15} \][/tex]
[tex]\[ -\frac{4\pi}{15} \approx -0.837 \quad \text{(which is negative and not in the valid range).} \][/tex]
2. [tex]\( k = 1 \)[/tex]:
[tex]\[ x = \pi - \frac{4\pi}{15} = \frac{11\pi}{15} \][/tex]
[tex]\[ \frac{11\pi}{15} \approx 2.303 \quad \text{radians (which is greater than } \frac{\pi}{18} \text{ and less than } 2\pi\text{, so it meets the condition).} \][/tex]
Summary of valid solutions within the given constraints:
[tex]\[ x = \frac{11\pi}{15} \][/tex]
So, the solutions to [tex]\( \cos \left(2x + \frac{\pi}{3}\right) = \cos \frac{\pi}{5} \)[/tex] within the range [tex]\( 10^\circ \leq x \leq 2\pi \)[/tex] are:
[tex]\[ x = \frac{11\pi}{15} \][/tex]
We begin by using the property of the cosine function that states if [tex]\( \cos A = \cos B \)[/tex], then [tex]\( A = 2k\pi \pm B \)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex].
For our equation, we set:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi \pm \frac{\pi}{5} \][/tex]
Let's first solve using the positive case:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi + \frac{\pi}{5} \][/tex]
Reorganize to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 2k\pi + \frac{\pi}{5} - \frac{\pi}{3} \][/tex]
[tex]\[ 2x = 2k\pi + \frac{\pi}{15} \left(5 - 3\right) \][/tex]
[tex]\[ 2x = 2k\pi + \frac{2\pi}{15} \][/tex]
[tex]\[ x = k\pi + \frac{\pi}{15} \][/tex]
Now, let's solve using the negative case:
[tex]\[ 2x + \frac{\pi}{3} = 2k\pi - \frac{\pi}{5} \][/tex]
Reorganize to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 2k\pi - \frac{\pi}{5} - \frac{\pi}{3} \][/tex]
[tex]\[ 2x = 2k\pi - \frac{\pi}{15} \left(5 + 3\right) \][/tex]
[tex]\[ 2x = 2k\pi - \frac{8\pi}{15} \][/tex]
[tex]\[ x = k\pi - \frac{4\pi}{15} \][/tex]
Next, we consider the constraint [tex]\( 10^\circ \leqslant x \leqslant 2\pi \)[/tex], which translates [tex]\( 10^\circ \)[/tex] to radians:
[tex]\[ 10^\circ = \frac{10\pi}{180} = \frac{\pi}{18} \][/tex]
The constraint becomes:
[tex]\[ \frac{\pi}{18} \leq x \leq 2\pi \][/tex]
We now solve for [tex]\( x \)[/tex] within this range.
For [tex]\( x = k\pi + \frac{\pi}{15} \)[/tex] we check if the possible [tex]\( x \)[/tex] values meet the constraint:
1. [tex]\( k = 0 \)[/tex]:
[tex]\[ x = \frac{\pi}{15} \][/tex]
[tex]\[ \frac{\pi}{15} \approx 0.209 \quad \text{radians (which is less than } \frac{\pi}{18}\text{), so it's not in the valid range.} \][/tex]
2. [tex]\( k = 1 \)[/tex]:
[tex]\[ x = \pi + \frac{\pi}{15} = \frac{16\pi}{15} \][/tex]
[tex]\[ \frac{16\pi}{15} \approx 3.351 \quad \text{radians (which is greater than } \frac{\pi}{18} \text{ and less than } 2\pi\text{, so it meets the condition.} \][/tex]
For [tex]\( x = k\pi - \frac{4\pi}{15} \)[/tex], check the possible [tex]\( x \)[/tex] values:
1. [tex]\( k = 0 \)[/tex]:
[tex]\[ x = -\frac{4\pi}{15} \][/tex]
[tex]\[ -\frac{4\pi}{15} \approx -0.837 \quad \text{(which is negative and not in the valid range).} \][/tex]
2. [tex]\( k = 1 \)[/tex]:
[tex]\[ x = \pi - \frac{4\pi}{15} = \frac{11\pi}{15} \][/tex]
[tex]\[ \frac{11\pi}{15} \approx 2.303 \quad \text{radians (which is greater than } \frac{\pi}{18} \text{ and less than } 2\pi\text{, so it meets the condition).} \][/tex]
Summary of valid solutions within the given constraints:
[tex]\[ x = \frac{11\pi}{15} \][/tex]
So, the solutions to [tex]\( \cos \left(2x + \frac{\pi}{3}\right) = \cos \frac{\pi}{5} \)[/tex] within the range [tex]\( 10^\circ \leq x \leq 2\pi \)[/tex] are:
[tex]\[ x = \frac{11\pi}{15} \][/tex]
