Answer :
Certainly! Let's solve the equation step by step.
Given the equation:
[tex]\[ (\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \][/tex]
We can start by using trigonometric identities. Recall the product-to-sum identities:
[tex]\[ \cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)] \][/tex]
and
[tex]\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \][/tex]
Using [tex]\(A = 3\theta\)[/tex] and [tex]\(B = \theta\)[/tex], we have:
[tex]\[ \cos 3\theta \cos \theta = \frac{1}{2} [\cos (3\theta + \theta) + \cos (3\theta - \theta)] = \frac{1}{2} [\cos 4\theta + \cos 2\theta] \][/tex]
and
[tex]\[ \sin 3\theta \sin \theta = \frac{1}{2} [\cos (3\theta - \theta) - \cos (3\theta + \theta)] = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Substitute these into the given equation:
[tex]\[ \frac{1}{2} [\cos 4\theta + \cos 2\theta] + 1 = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Multiply through by 2 to clear the fractions:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Rearrange terms to combine like terms:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Combine the [tex]\(\cos 4\theta\)[/tex] terms:
[tex]\[ 2\cos 4\theta + 2 = 0 \][/tex]
Subtract 2 from both sides:
[tex]\[ 2\cos 4\theta = -2 \][/tex]
Divide by 2:
[tex]\[ \cos 4\theta = -1 \][/tex]
The general solution to [tex]\(\cos (x) = -1\)[/tex] is:
[tex]\[ x = \pi + 2k\pi \quad \text{for integer } k \][/tex]
Set [tex]\(x = 4\theta\)[/tex]:
[tex]\[ 4\theta = \pi + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{k\pi}{2} \][/tex]
Now, let's find the specific values of [tex]\(\theta\)[/tex] that satisfy the original equation within one full cycle (0 to [tex]\(2\pi\)[/tex]). We consider the first few values of [tex]\(k\)[/tex]:
For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \][/tex]
For [tex]\(k = -1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \][/tex]
For [tex]\(k = -2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \][/tex]
Thus, the solutions to the equation [tex]\((\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \)[/tex] are:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
Given the equation:
[tex]\[ (\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \][/tex]
We can start by using trigonometric identities. Recall the product-to-sum identities:
[tex]\[ \cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)] \][/tex]
and
[tex]\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \][/tex]
Using [tex]\(A = 3\theta\)[/tex] and [tex]\(B = \theta\)[/tex], we have:
[tex]\[ \cos 3\theta \cos \theta = \frac{1}{2} [\cos (3\theta + \theta) + \cos (3\theta - \theta)] = \frac{1}{2} [\cos 4\theta + \cos 2\theta] \][/tex]
and
[tex]\[ \sin 3\theta \sin \theta = \frac{1}{2} [\cos (3\theta - \theta) - \cos (3\theta + \theta)] = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Substitute these into the given equation:
[tex]\[ \frac{1}{2} [\cos 4\theta + \cos 2\theta] + 1 = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Multiply through by 2 to clear the fractions:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Rearrange terms to combine like terms:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Combine the [tex]\(\cos 4\theta\)[/tex] terms:
[tex]\[ 2\cos 4\theta + 2 = 0 \][/tex]
Subtract 2 from both sides:
[tex]\[ 2\cos 4\theta = -2 \][/tex]
Divide by 2:
[tex]\[ \cos 4\theta = -1 \][/tex]
The general solution to [tex]\(\cos (x) = -1\)[/tex] is:
[tex]\[ x = \pi + 2k\pi \quad \text{for integer } k \][/tex]
Set [tex]\(x = 4\theta\)[/tex]:
[tex]\[ 4\theta = \pi + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{k\pi}{2} \][/tex]
Now, let's find the specific values of [tex]\(\theta\)[/tex] that satisfy the original equation within one full cycle (0 to [tex]\(2\pi\)[/tex]). We consider the first few values of [tex]\(k\)[/tex]:
For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \][/tex]
For [tex]\(k = -1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \][/tex]
For [tex]\(k = -2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \][/tex]
Thus, the solutions to the equation [tex]\((\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \)[/tex] are:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
