Question

If \(cos(pθ)+cos(qθ) = 0\) and \(p\neq q\), then the general value of \(θ\) (where n is any integer) is...

• A. \(\frac{(3n+1)π}{p-q}\)
• B. \(\frac{(2n+1)π}{p\pm q}\)
• C. \(\frac{(n\pm 1)π}{p\pm q}\)
• D. \(\frac{(n+2)π}{p+q}\)

Hint:

Use the sum-to-product formula and set each cosine to zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Given \(\cos p\theta+\cos q\theta=0\), so \(\cos p\theta=-\cos q\theta=\cos(\pi-q\theta)\). If \(\cos\alpha=\cos\beta\) then \(\alpha=2n\pi\pm\beta\).

Step 2: Apply:
\[ p\theta=2n\pi\pm(\pi-q\theta) \]

Step 3: Take the plus sign:
\(p\theta=2n\pi+\pi-q\theta\), so \((p+q)\theta=(2n+1)\pi\) and \(\theta=\dfrac{(2n+1)\pi}{p+q}\).

Step 4: Take the minus sign:
\(p\theta=2n\pi-\pi+q\theta\), so \((p-q)\theta=(2n-1)\pi\) and \(\theta=\dfrac{(2n-1)\pi}{p-q}\). As \(n\) runs over all integers, \(2n-1\) covers all odd numbers, the same as \(2n+1\).

Step 5: Combine:
\(\theta=\dfrac{(2n+1)\pi}{p\pm q}\), option (B).

Final Answer:
The general solution is (2n+1) pi/(p +- q). \[ \boxed{\theta=\frac{(2n+1)\pi}{p\pm q}} \]