Find all solutions to the equation.
$\(\left\)(\(\cos\[\theta\)+\(\sin\]\theta\[\right\))\(\left\)(\(\cos\]\theta\)-\(\sin\[\theta\]\right\))=-\(\frac\)12$

• A. $\(\theta\)=\(\frac{5\pi}{12}\)+2\(\pi\) n,\(\frac{7\pi}{12}\)+2\(\pi\) n$
• B. $\(\theta\)=\(\frac{2\pi}{3}\)+2\(\pi\) n,\(\frac{4\pi}{3}\)+2\(\pi\) n$
• C. $\(\theta\)=\(\frac{\pi}{3}\)+2\(\pi\) n,\(\frac{2\pi}{3}\)+2\(\pi\) n$
• D. $\(\theta\)=\(\frac{\pi}{3}\)+\(\pi\) n,\(\frac{2\pi}{3}\)+\(\pi\) n$

Answer: (D)

Verified step by step guidance

1. Start by expanding the expression \((\cos\theta + \sin\theta)(\cos\theta - \sin\theta)\). This is a difference of squares, which can be simplified using the identity \(a^2 - b^2 = (a+b)(a-b)\).
2. Apply the identity: \((\cos\theta + \sin\theta)(\cos\theta - \sin\theta) = \cos^2\theta - \sin^2\theta\).
3. Recognize that \(\cos^2\theta - \sin^2\theta\) can be rewritten using the identity \(\cos^2\theta - \sin^2\theta = \cos(2\theta)\).
4. Set the equation \(\cos(2\theta) = -\frac{1}{2}\) and solve for \(2\theta\). The solutions for \(\cos(\alpha) = -\frac{1}{2}\) are \(\alpha = \frac{2\pi}{3} + 2\pi n\) and \(\alpha = \frac{4\pi}{3} + 2\pi n\).
5. Divide the solutions for \(2\theta\) by 2 to find \(\theta\). This gives \(\theta = \frac{\pi}{3} + \pi n\) and \(\theta = \frac{2\pi}{3} + \pi n\). These are the solutions to the original equation.